5: Continuous Random Variables
Spring 2021
Matthew Blackwell
Gov 2002 (Harvard)
1 / 16 • How to characterize uncertainty about data that takes on discrete values.
• Learned how to define distributions (p.m.f., c.d.f.) and how to summarize.
• Now: define the same ideas for r.v.s that can take on any real value.
Where are we? Where are we going?
• Last few weeks: discrete random variables.
2 / 16 • Learned how to define distributions (p.m.f., c.d.f.) and how to summarize.
• Now: define the same ideas for r.v.s that can take on any real value.
Where are we? Where are we going?
• Last few weeks: discrete random variables.
• How to characterize uncertainty about data that takes on discrete values.
2 / 16 • Now: define the same ideas for r.v.s that can take on any real value.
Where are we? Where are we going?
• Last few weeks: discrete random variables.
• How to characterize uncertainty about data that takes on discrete values.
• Learned how to define distributions (p.m.f., c.d.f.) and how to summarize.
2 / 16 Where are we? Where are we going?
• Last few weeks: discrete random variables.
• How to characterize uncertainty about data that takes on discrete values.
• Learned how to define distributions (p.m.f., c.d.f.) and how to summarize.
• Now: define the same ideas for r.v.s that can take on any real value.
2 / 16 • Suppose ℙ(혟 = 혹) = 휀 for 혹 ∈ (ퟢ, ퟣ) where 휀 is a very small number. • What’s the probability of being between 0 and 1? • There are an infinite number of real numbers between 0 and 1:
ퟢ.ퟤퟥퟤퟪퟩퟫퟪퟩퟥ … ퟢ.ퟧퟩퟤퟨퟥퟢퟦퟪퟩퟦퟥ … ퟢ.ퟫퟪퟤퟥퟨퟣퟤퟫퟪퟦ …
• Each one has probability 휀 ⇝ ℙ(혟 ∈ (ퟢ, ퟣ)) = ∞ × 휀 = ∞
• What if 혟 can take any value on any real value?
• Can we just specify ℙ(혟 = 혹) for all 혹?
• No! Proof by counterexample:
• But ℙ(혟 ∈ (ퟢ, ퟣ)) must be less than 1! ⇝ ℙ(혟 = 혹) must be 0.
Continuous r.v.s
• Discrete r.v.: specify ℙ(혟 = 혹) for all possible values ⇝ p.m.f.
3 / 16 • Suppose ℙ(혟 = 혹) = 휀 for 혹 ∈ (ퟢ, ퟣ) where 휀 is a very small number. • What’s the probability of being between 0 and 1? • There are an infinite number of real numbers between 0 and 1:
ퟢ.ퟤퟥퟤퟪퟩퟫퟪퟩퟥ … ퟢ.ퟧퟩퟤퟨퟥퟢퟦퟪퟩퟦퟥ … ퟢ.ퟫퟪퟤퟥퟨퟣퟤퟫퟪퟦ …
• Each one has probability 휀 ⇝ ℙ(혟 ∈ (ퟢ, ퟣ)) = ∞ × 휀 = ∞
• Can we just specify ℙ(혟 = 혹) for all 혹?
• No! Proof by counterexample:
• But ℙ(혟 ∈ (ퟢ, ퟣ)) must be less than 1! ⇝ ℙ(혟 = 혹) must be 0.
Continuous r.v.s
• Discrete r.v.: specify ℙ(혟 = 혹) for all possible values ⇝ p.m.f.
• What if 혟 can take any value on any real value?
3 / 16 • Suppose ℙ(혟 = 혹) = 휀 for 혹 ∈ (ퟢ, ퟣ) where 휀 is a very small number. • What’s the probability of being between 0 and 1? • There are an infinite number of real numbers between 0 and 1:
ퟢ.ퟤퟥퟤퟪퟩퟫퟪퟩퟥ … ퟢ.ퟧퟩퟤퟨퟥퟢퟦퟪퟩퟦퟥ … ퟢ.ퟫퟪퟤퟥퟨퟣퟤퟫퟪퟦ …
• Each one has probability 휀 ⇝ ℙ(혟 ∈ (ퟢ, ퟣ)) = ∞ × 휀 = ∞
• No! Proof by counterexample:
• But ℙ(혟 ∈ (ퟢ, ퟣ)) must be less than 1! ⇝ ℙ(혟 = 혹) must be 0.
Continuous r.v.s
• Discrete r.v.: specify ℙ(혟 = 혹) for all possible values ⇝ p.m.f.
• What if 혟 can take any value on any real value?
• Can we just specify ℙ(혟 = 혹) for all 혹?
3 / 16 • Suppose ℙ(혟 = 혹) = 휀 for 혹 ∈ (ퟢ, ퟣ) where 휀 is a very small number. • What’s the probability of being between 0 and 1? • There are an infinite number of real numbers between 0 and 1:
ퟢ.ퟤퟥퟤퟪퟩퟫퟪퟩퟥ … ퟢ.ퟧퟩퟤퟨퟥퟢퟦퟪퟩퟦퟥ … ퟢ.ퟫퟪퟤퟥퟨퟣퟤퟫퟪퟦ …
• Each one has probability 휀 ⇝ ℙ(혟 ∈ (ퟢ, ퟣ)) = ∞ × 휀 = ∞
• But ℙ(혟 ∈ (ퟢ, ퟣ)) must be less than 1! ⇝ ℙ(혟 = 혹) must be 0.
Continuous r.v.s
• Discrete r.v.: specify ℙ(혟 = 혹) for all possible values ⇝ p.m.f.
• What if 혟 can take any value on any real value?
• Can we just specify ℙ(혟 = 혹) for all 혹?
• No! Proof by counterexample:
3 / 16 • What’s the probability of being between 0 and 1? • There are an infinite number of real numbers between 0 and 1:
ퟢ.ퟤퟥퟤퟪퟩퟫퟪퟩퟥ … ퟢ.ퟧퟩퟤퟨퟥퟢퟦퟪퟩퟦퟥ … ퟢ.ퟫퟪퟤퟥퟨퟣퟤퟫퟪퟦ …
• Each one has probability 휀 ⇝ ℙ(혟 ∈ (ퟢ, ퟣ)) = ∞ × 휀 = ∞
• But ℙ(혟 ∈ (ퟢ, ퟣ)) must be less than 1! ⇝ ℙ(혟 = 혹) must be 0.
Continuous r.v.s
• Discrete r.v.: specify ℙ(혟 = 혹) for all possible values ⇝ p.m.f.
• What if 혟 can take any value on any real value?
• Can we just specify ℙ(혟 = 혹) for all 혹?
• No! Proof by counterexample:
• Suppose ℙ(혟 = 혹) = 휀 for 혹 ∈ (ퟢ, ퟣ) where 휀 is a very small number.
3 / 16 • There are an infinite number of real numbers between 0 and 1:
ퟢ.ퟤퟥퟤퟪퟩퟫퟪퟩퟥ … ퟢ.ퟧퟩퟤퟨퟥퟢퟦퟪퟩퟦퟥ … ퟢ.ퟫퟪퟤퟥퟨퟣퟤퟫퟪퟦ …
• Each one has probability 휀 ⇝ ℙ(혟 ∈ (ퟢ, ퟣ)) = ∞ × 휀 = ∞
• But ℙ(혟 ∈ (ퟢ, ퟣ)) must be less than 1! ⇝ ℙ(혟 = 혹) must be 0.
Continuous r.v.s
• Discrete r.v.: specify ℙ(혟 = 혹) for all possible values ⇝ p.m.f.
• What if 혟 can take any value on any real value?
• Can we just specify ℙ(혟 = 혹) for all 혹?
• No! Proof by counterexample:
• Suppose ℙ(혟 = 혹) = 휀 for 혹 ∈ (ퟢ, ퟣ) where 휀 is a very small number. • What’s the probability of being between 0 and 1?
3 / 16 • Each one has probability 휀 ⇝ ℙ(혟 ∈ (ퟢ, ퟣ)) = ∞ × 휀 = ∞
• But ℙ(혟 ∈ (ퟢ, ퟣ)) must be less than 1! ⇝ ℙ(혟 = 혹) must be 0.
Continuous r.v.s
• Discrete r.v.: specify ℙ(혟 = 혹) for all possible values ⇝ p.m.f.
• What if 혟 can take any value on any real value?
• Can we just specify ℙ(혟 = 혹) for all 혹?
• No! Proof by counterexample:
• Suppose ℙ(혟 = 혹) = 휀 for 혹 ∈ (ퟢ, ퟣ) where 휀 is a very small number. • What’s the probability of being between 0 and 1? • There are an infinite number of real numbers between 0 and 1:
ퟢ.ퟤퟥퟤퟪퟩퟫퟪퟩퟥ … ퟢ.ퟧퟩퟤퟨퟥퟢퟦퟪퟩퟦퟥ … ퟢ.ퟫퟪퟤퟥퟨퟣퟤퟫퟪퟦ …
3 / 16 • But ℙ(혟 ∈ (ퟢ, ퟣ)) must be less than 1! ⇝ ℙ(혟 = 혹) must be 0.
Continuous r.v.s
• Discrete r.v.: specify ℙ(혟 = 혹) for all possible values ⇝ p.m.f.
• What if 혟 can take any value on any real value?
• Can we just specify ℙ(혟 = 혹) for all 혹?
• No! Proof by counterexample:
• Suppose ℙ(혟 = 혹) = 휀 for 혹 ∈ (ퟢ, ퟣ) where 휀 is a very small number. • What’s the probability of being between 0 and 1? • There are an infinite number of real numbers between 0 and 1:
ퟢ.ퟤퟥퟤퟪퟩퟫퟪퟩퟥ … ퟢ.ퟧퟩퟤퟨퟥퟢퟦퟪퟩퟦퟥ … ퟢ.ퟫퟪퟤퟥퟨퟣퟤퟫퟪퟦ …
• Each one has probability 휀 ⇝ ℙ(혟 ∈ (ퟢ, ퟣ)) = ∞ × 휀 = ∞
3 / 16 Continuous r.v.s
• Discrete r.v.: specify ℙ(혟 = 혹) for all possible values ⇝ p.m.f.
• What if 혟 can take any value on any real value?
• Can we just specify ℙ(혟 = 혹) for all 혹?
• No! Proof by counterexample:
• Suppose ℙ(혟 = 혹) = 휀 for 혹 ∈ (ퟢ, ퟣ) where 휀 is a very small number. • What’s the probability of being between 0 and 1? • There are an infinite number of real numbers between 0 and 1:
ퟢ.ퟤퟥퟤퟪퟩퟫퟪퟩퟥ … ퟢ.ퟧퟩퟤퟨퟥퟢퟦퟪퟩퟦퟥ … ퟢ.ퟫퟪퟤퟥퟨퟣퟤퟫퟪퟦ …
• Each one has probability 휀 ⇝ ℙ(혟 ∈ (ퟢ, ퟣ)) = ∞ × 휀 = ∞
• But ℙ(혟 ∈ (ퟢ, ퟣ)) must be less than 1! ⇝ ℙ(혟 = 혹) must be 0.
3 / 16 3.1415926535897932384626433832795028841971693993751058209749445923078164
0628620899862803482534211706798214808651328230664709384460955058223172
5359408128481117450284102701938521105559644622948954930381964428810975
6659334461284756482337867831652712019091456485669234603486104543266482
1339360726024914127372458700660631558817488152092096282925409171536436
7892590360011330530548820466521384146951941511609433057270365759591953
0921861173819326117931051185480744623799627495673518857527248912279381
8301194912983367336244065664308602139494639522473719070217986094370277
0539217176293176752384674818467669405132000568127145263560827785771342
7577896091736371787214684409012249534301465495853710507922796892589235
4201995611212902196086403441815981362977477130996051870721134999999837
2978049951059731732816096318595024459455346908302642522308253344685035
2619311881710100031378387528865875332083814206171776691473035982534904
2875546873115956286388235378759375195778185778053217122680661300192787
6611195909216420198938095257201065485863278865936153381827968230301952
0353018529689957736225994138912497217752834791315155748572424541506959
5082953311686172785588907509838175463746493931925506040092770167113900
984882401285836160356370766010471018194295559619894676783744...
Thought experiment: draw a random real value between 0 and 10. What’s the probability that we draw a value that is exact equal to 휋?
4 / 16 Thought experiment: draw a random real value between 0 and 10. What’s the probability that we draw a value that is exact equal to 휋?
3.1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164
0628620899 8628034825 3421170679 8214808651 3282306647 0938446095 5058223172
5359408128 4811174502 8410270193 8521105559 6446229489 5493038196 4428810975
6659334461 2847564823 3786783165 2712019091 4564856692 3460348610 4543266482
1339360726 0249141273 7245870066 0631558817 4881520920 9628292540 9171536436
7892590360 0113305305 4882046652 1384146951 9415116094 3305727036 5759591953
0921861173 8193261179 3105118548 0744623799 6274956735 1885752724 8912279381
8301194912 9833673362 4406566430 8602139494 6395224737 1907021798 6094370277
0539217176 2931767523 8467481846 7669405132 0005681271 4526356082 7785771342
7577896091 7363717872 1468440901 2249534301 4654958537 1050792279 6892589235
4201995611 2129021960 8640344181 5981362977 4771309960 5187072113 4999999837
2978049951 0597317328 1609631859 5024459455 3469083026 4252230825 3344685035
2619311881 7101000313 7838752886 5875332083 8142061717 7669147303 5982534904
2875546873 1159562863 8823537875 9375195778 1857780532 1712268066 1300192787
6611195909 2164201989 3809525720 1065485863 2788659361 5338182796 8230301952
0353018529 6899577362 2599413891 2497217752 8347913151 5574857242 4541506959
5082953311 6861727855 8890750983 8175463746 4939319255 0604009277 0167113900 4 / 16 9848824012 8583616035 6370766010 4710181942 9555961989 4676783744... • Support of 혟 is all values such that 혧혟 (혹) > ퟢ.
• Intuition of a density:
혧 (혹ퟢ)휀 ≈ ℙ(혟 ∈ (혹ퟢ − 휀/ퟤ, 혹ퟢ + 휀/ퟤ))
혤 ℙ(혟 = 혤) = ∫ 혧 (혹)혥혹 = ퟢ • Probability of a point mass: 혤 혟
Probability density functions
Definition
A r.v., 혟, is continuous if there exists a nonnegative function on ℝ, 혧혟 called the probability density function (p.d.f.) such that for any interval, (혢, 혣):
혣
ℙ(혢 ≤ 혟 ≤ 혣) = ∫ 혧혟 (혹)혥혹 혢
• ⇝ the probability of a region is the area under the p.d.f. for that region.
5 / 16 • Intuition of a density:
혧 (혹ퟢ)휀 ≈ ℙ(혟 ∈ (혹ퟢ − 휀/ퟤ, 혹ퟢ + 휀/ퟤ))
혤 ℙ(혟 = 혤) = ∫ 혧 (혹)혥혹 = ퟢ • Probability of a point mass: 혤 혟
Probability density functions
Definition
A r.v., 혟, is continuous if there exists a nonnegative function on ℝ, 혧혟 called the probability density function (p.d.f.) such that for any interval, (혢, 혣):
혣
ℙ(혢 ≤ 혟 ≤ 혣) = ∫ 혧혟 (혹)혥혹 혢
• ⇝ the probability of a region is the area under the p.d.f. for that region.
• Support of 혟 is all values such that 혧혟 (혹) > ퟢ.
5 / 16 혤 ℙ(혟 = 혤) = ∫ 혧 (혹)혥혹 = ퟢ • Probability of a point mass: 혤 혟
Probability density functions
Definition
A r.v., 혟, is continuous if there exists a nonnegative function on ℝ, 혧혟 called the probability density function (p.d.f.) such that for any interval, (혢, 혣):
혣
ℙ(혢 ≤ 혟 ≤ 혣) = ∫ 혧혟 (혹)혥혹 혢
• ⇝ the probability of a region is the area under the p.d.f. for that region.
• Support of 혟 is all values such that 혧혟 (혹) > ퟢ.
• Intuition of a density:
혧 (혹ퟢ)휀 ≈ ℙ(혟 ∈ (혹ퟢ − 휀/ퟤ, 혹ퟢ + 휀/ퟤ))
5 / 16 Probability density functions
Definition
A r.v., 혟, is continuous if there exists a nonnegative function on ℝ, 혧혟 called the probability density function (p.d.f.) such that for any interval, (혢, 혣):
혣
ℙ(혢 ≤ 혟 ≤ 혣) = ∫ 혧혟 (혹)혥혹 혢
• ⇝ the probability of a region is the area under the p.d.f. for that region.
• Support of 혟 is all values such that 혧혟 (혹) > ퟢ.
• Intuition of a density:
혧 (혹ퟢ)휀 ≈ ℙ(혟 ∈ (혹ퟢ − 휀/ퟤ, 혹ퟢ + 휀/ퟤ))
혤 ℙ(혟 = 혤) = ∫ 혧 (혹)혥혹 = ퟢ • Probability of a point mass: 혤 혟
5 / 16 • Nonnegative: 혧혟 (혹) > ퟢ ∞ ∫ 혧 (혹)혥혹 = ퟣ • Integrates to 1: −∞ 혟
The p.d.f.
Logistic distribution (p.d.f.) 0.4
0.3
0.2 P(0 0.1 0.0 -4 -2 0 2 4 x • Properties of a valid p.d.f.: 6 / 16 ∞ ∫ 혧 (혹)혥혹 = ퟣ • Integrates to 1: −∞ 혟 The p.d.f. Logistic distribution (p.d.f.) 0.4 0.3 0.2 P(0 0.1 0.0 -4 -2 0 2 4 x • Properties of a valid p.d.f.: • Nonnegative: 혧혟 (혹) > ퟢ 6 / 16 The p.d.f. Logistic distribution (p.d.f.) 0.4 0.3 0.2 P(0 0.1 0.0 -4 -2 0 2 4 x • Properties of a valid p.d.f.: • Nonnegative: 혧혟 (혹) > ퟢ ∞ ∫ 혧 (혹)혥혹 = ퟣ • Integrates to 1: −∞ 혟 6 / 16 • ℙ(혢 < 혟 < 혣) = ℙ(혢 < 혟 ≤ 혣) = ℙ(혢 ≤ 혟 < 혣) = ℙ(혢 ≤ 혟 ≤ 혣). • With continuous we don’t have to worry about < vs ≤. Cumulative distribution functions Continuous r.v. c.d.f. The cumulative distribution function of a continuous r.v. 혟 is given by 혹 혍혟 (혹) ≡ ℙ(혟 ≤ 혹) = ∫ 혧혟 (혵)혥혵. −∞ • By the fundamental theorem of calculus: 혣 ℙ(혢 < 혟 ≤ 혣) = 혍 (혣) − 혍 (혢) = ∫ 혧 (혹)혥혹. 혢 7 / 16 • ℙ(혢 < 혟 < 혣) = ℙ(혢 < 혟 ≤ 혣) = ℙ(혢 ≤ 혟 < 혣) = ℙ(혢 ≤ 혟 ≤ 혣). Cumulative distribution functions Continuous r.v. c.d.f. The cumulative distribution function of a continuous r.v. 혟 is given by 혹 혍혟 (혹) ≡ ℙ(혟 ≤ 혹) = ∫ 혧혟 (혵)혥혵. −∞ • By the fundamental theorem of calculus: 혣 ℙ(혢 < 혟 ≤ 혣) = 혍 (혣) − 혍 (혢) = ∫ 혧 (혹)혥혹. 혢 • With continuous we don’t have to worry about < vs ≤. 7 / 16 Cumulative distribution functions Continuous r.v. c.d.f. The cumulative distribution function of a continuous r.v. 혟 is given by 혹 혍혟 (혹) ≡ ℙ(혟 ≤ 혹) = ∫ 혧혟 (혵)혥혵. −∞ • By the fundamental theorem of calculus: 혣 ℙ(혢 < 혟 ≤ 혣) = 혍 (혣) − 혍 (혢) = ∫ 혧 (혹)혥혹. 혢 • With continuous we don’t have to worry about < vs ≤. • ℙ(혢 < 혟 < 혣) = ℙ(혢 < 혟 ≤ 혣) = ℙ(혢 ≤ 혟 < 혣) = ℙ(혢 ≤ 혟 ≤ 혣). 7 / 16 Continuous c.d.f. Logistic p.d.f. Logistic c.d.f. 0.5 1.0 0.4 0.8 0.3 0.6 f(x) F(x) 0.2 0.4 0.1 0.2 0.0 0.0 -4 -2 0 2 4 -4 -2 0 2 4 x x • c.d.f. for continuous r.v. = integral of p.d.f. up to a certain value. 8 / 16 • If (혤, 혥) is a subinterval of (혢, 혣) then ℙ(혜 ∈ (혤, 혥)) is proportional to 혤 − 혥 • Distribution of 혜 conditional on being in (혤, 혥) is Unif(혤, 혥). • Intuitively, every equal-sized interval has the same probability. • How can figure out the p.d.f. for such a distribution? Definition A continuous r.v. 혜 has a Uniform distribution on the interval (혢, 혣) if its p.d.f. is {⎧ ퟣ for 혹 ∈ [혢, 혣] 혧 (혹) = 혣−혢 ⎨ ⎩{ퟢ otherwise Continuous uniform distribution • Simple and really important continuous distribution: uniform. 9 / 16 • If (혤, 혥) is a subinterval of (혢, 혣) then ℙ(혜 ∈ (혤, 혥)) is proportional to 혤 − 혥 • Distribution of 혜 conditional on being in (혤, 혥) is Unif(혤, 혥). • How can figure out the p.d.f. for such a distribution? Definition A continuous r.v. 혜 has a Uniform distribution on the interval (혢, 혣) if its p.d.f. is {⎧ ퟣ for 혹 ∈ [혢, 혣] 혧 (혹) = 혣−혢 ⎨ ⎩{ퟢ otherwise Continuous uniform distribution • Simple and really important continuous distribution: uniform. • Intuitively, every equal-sized interval has the same probability. 9 / 16 • If (혤, 혥) is a subinterval of (혢, 혣) then ℙ(혜 ∈ (혤, 혥)) is proportional to 혤 − 혥 • Distribution of 혜 conditional on being in (혤, 혥) is Unif(혤, 혥). Definition A continuous r.v. 혜 has a Uniform distribution on the interval (혢, 혣) if its p.d.f. is {⎧ ퟣ for 혹 ∈ [혢, 혣] 혧 (혹) = 혣−혢 ⎨ ⎩{ퟢ otherwise Continuous uniform distribution • Simple and really important continuous distribution: uniform. • Intuitively, every equal-sized interval has the same probability. • How can figure out the p.d.f. for such a distribution? 9 / 16 • If (혤, 혥) is a subinterval of (혢, 혣) then ℙ(혜 ∈ (혤, 혥)) is proportional to 혤 − 혥 • Distribution of 혜 conditional on being in (혤, 혥) is Unif(혤, 혥). Continuous uniform distribution • Simple and really important continuous distribution: uniform. • Intuitively, every equal-sized interval has the same probability. • How can figure out the p.d.f. for such a distribution? Definition A continuous r.v. 혜 has a Uniform distribution on the interval (혢, 혣) if its p.d.f. is {⎧ ퟣ for 혹 ∈ [혢, 혣] 혧 (혹) = 혣−혢 ⎨ ⎩{ퟢ otherwise 9 / 16 • Distribution of 혜 conditional on being in (혤, 혥) is Unif(혤, 혥). Continuous uniform distribution • Simple and really important continuous distribution: uniform. • Intuitively, every equal-sized interval has the same probability. • How can figure out the p.d.f. for such a distribution? Definition A continuous r.v. 혜 has a Uniform distribution on the interval (혢, 혣) if its p.d.f. is {⎧ ퟣ for 혹 ∈ [혢, 혣] 혧 (혹) = 혣−혢 ⎨ ⎩{ퟢ otherwise • If (혤, 혥) is a subinterval of (혢, 혣) then ℙ(혜 ∈ (혤, 혥)) is proportional to 혤 − 혥 9 / 16 Continuous uniform distribution • Simple and really important continuous distribution: uniform. • Intuitively, every equal-sized interval has the same probability. • How can figure out the p.d.f. for such a distribution? Definition A continuous r.v. 혜 has a Uniform distribution on the interval (혢, 혣) if its p.d.f. is {⎧ ퟣ for 혹 ∈ [혢, 혣] 혧 (혹) = 혣−혢 ⎨ ⎩{ퟢ otherwise • If (혤, 혥) is a subinterval of (혢, 혣) then ℙ(혜 ∈ (혤, 혥)) is proportional to 혤 − 혥 • Distribution of 혜 conditional on being in (혤, 혥) is Unif(혤, 혥). 9 / 16 • Linear transformations of uniforms preserve the uniform distribution. • Location-scale transformation: Let 혜 ∼ Unif(혢, 혣). Then 혜̃ = 혤혜 + 혥 is Unif(혤혢 + 혥, 혤혣 + 혥) Uniform pdf and cdf 1.2 1.0 1.0 0.8 0.8 0.6 f(x) 0.6 F(x) 0.4 0.4 0.2 0.2 0.0 0.0 0.0 0.5 1.0 0.0 0.5 1.0 x x 10 / 16 • Linear transformations of uniforms preserve the uniform distribution. Uniform pdf and cdf 1.2 1.0 1.0 0.8 0.8 0.6 f(x) 0.6 F(x) 0.4 0.4 0.2 0.2 0.0 0.0 0.0 0.5 1.0 0.0 0.5 1.0 x x • Location-scale transformation: Let 혜 ∼ Unif(혢, 혣). Then 혜̃ = 혤혜 + 혥 is Unif(혤혢 + 혥, 혤혣 + 혥) 10 / 16 Uniform pdf and cdf 1.2 1.0 1.0 0.8 0.8 0.6 f(x) 0.6 F(x) 0.4 0.4 0.2 0.2 0.0 0.0 0.0 0.5 1.0 0.0 0.5 1.0 x x • Location-scale transformation: Let 혜 ∼ Unif(혢, 혣). Then 혜̃ = 혤혜 + 혥 is Unif(혤혢 + 혥, 혤혣 + 혥) • Linear transformations of uniforms preserve the uniform distribution. 10 / 16 • In particular, we still have 핍[혟] = 피[혟 ퟤ] − (피[혟])ퟤ 피[혜] = (혢 + 혣)/ퟤ ∞ 피[혨(혟)] = ∫ 혨(혹)혧 (혹)혥혹 • LOTUS with continuous r.v.s: −∞ 혟 • Variance of a continuous r.v.s: ∞ 핍[혟] = 피[(혟 − 피[혟])ퟤ] = ∫ (혹 − 피[혟])ퟤ혥혹 −∞ • Linearity and other properties of 피[] and 핍[] still hold! ∞ 피[혟] = ∫ 혹혥혍 (혹) • Unifying notation you may see: −∞ • Expectation of a uniform (0,1): Expectation for a continuous r.v. • Expectation of a continuous r.v.: ∞ 피[혟] = ∫ 혹혧혟 (혹)혥혹 −∞ 11 / 16 • In particular, we still have 핍[혟] = 피[혟 ퟤ] − (피[혟])ퟤ 피[혜] = (혢 + 혣)/ퟤ ∞ 피[혨(혟)] = ∫ 혨(혹)혧 (혹)혥혹 • LOTUS with continuous r.v.s: −∞ 혟 • Variance of a continuous r.v.s: ∞ 핍[혟] = 피[(혟 − 피[혟])ퟤ] = ∫ (혹 − 피[혟])ퟤ혥혹 −∞ • Linearity and other properties of 피[] and 핍[] still hold! • Expectation of a uniform (0,1): Expectation for a continuous r.v. • Expectation of a continuous r.v.: ∞ 피[혟] = ∫ 혹혧혟 (혹)혥혹 −∞ ∞ 피[혟] = ∫ 혹혥혍 (혹) • Unifying notation you may see: −∞ 11 / 16 • In particular, we still have 핍[혟] = 피[혟 ퟤ] − (피[혟])ퟤ ∞ 피[혨(혟)] = ∫ 혨(혹)혧 (혹)혥혹 • LOTUS with continuous r.v.s: −∞ 혟 • Variance of a continuous r.v.s: ∞ 핍[혟] = 피[(혟 − 피[혟])ퟤ] = ∫ (혹 − 피[혟])ퟤ혥혹 −∞ • Linearity and other properties of 피[] and 핍[] still hold! 피[혜] = (혢 + 혣)/ퟤ Expectation for a continuous r.v. • Expectation of a continuous r.v.: ∞ 피[혟] = ∫ 혹혧혟 (혹)혥혹 −∞ ∞ 피[혟] = ∫ 혹혥혍 (혹) • Unifying notation you may see: −∞ • Expectation of a uniform (0,1): 11 / 16 • In particular, we still have 핍[혟] = 피[혟 ퟤ] − (피[혟])ퟤ ∞ 피[혨(혟)] = ∫ 혨(혹)혧 (혹)혥혹 • LOTUS with continuous r.v.s: −∞ 혟 • Variance of a continuous r.v.s: ∞ 핍[혟] = 피[(혟 − 피[혟])ퟤ] = ∫ (혹 − 피[혟])ퟤ혥혹 −∞ • Linearity and other properties of 피[] and 핍[] still hold! Expectation for a continuous r.v. • Expectation of a continuous r.v.: ∞ 피[혟] = ∫ 혹혧혟 (혹)혥혹 −∞ ∞ 피[혟] = ∫ 혹혥혍 (혹) • Unifying notation you may see: −∞ • Expectation of a uniform (0,1): 피[혜] = (혢 + 혣)/ퟤ 11 / 16 • In particular, we still have 핍[혟] = 피[혟 ퟤ] − (피[혟])ퟤ • Variance of a continuous r.v.s: ∞ 핍[혟] = 피[(혟 − 피[혟])ퟤ] = ∫ (혹 − 피[혟])ퟤ혥혹 −∞ • Linearity and other properties of 피[] and 핍[] still hold! Expectation for a continuous r.v. • Expectation of a continuous r.v.: ∞ 피[혟] = ∫ 혹혧혟 (혹)혥혹 −∞ ∞ 피[혟] = ∫ 혹혥혍 (혹) • Unifying notation you may see: −∞ • Expectation of a uniform (0,1): 피[혜] = (혢 + 혣)/ퟤ ∞ 피[혨(혟)] = ∫ 혨(혹)혧 (혹)혥혹 • LOTUS with continuous r.v.s: −∞ 혟 11 / 16 • In particular, we still have 핍[혟] = 피[혟 ퟤ] − (피[혟])ퟤ • Linearity and other properties of 피[] and 핍[] still hold! Expectation for a continuous r.v. • Expectation of a continuous r.v.: ∞ 피[혟] = ∫ 혹혧혟 (혹)혥혹 −∞ ∞ 피[혟] = ∫ 혹혥혍 (혹) • Unifying notation you may see: −∞ • Expectation of a uniform (0,1): 피[혜] = (혢 + 혣)/ퟤ ∞ 피[혨(혟)] = ∫ 혨(혹)혧 (혹)혥혹 • LOTUS with continuous r.v.s: −∞ 혟 • Variance of a continuous r.v.s: ∞ 핍[혟] = 피[(혟 − 피[혟])ퟤ] = ∫ (혹 − 피[혟])ퟤ혥혹 −∞ 11 / 16 • In particular, we still have 핍[혟] = 피[혟 ퟤ] − (피[혟])ퟤ Expectation for a continuous r.v. • Expectation of a continuous r.v.: ∞ 피[혟] = ∫ 혹혧혟 (혹)혥혹 −∞ ∞ 피[혟] = ∫ 혹혥혍 (혹) • Unifying notation you may see: −∞ • Expectation of a uniform (0,1): 피[혜] = (혢 + 혣)/ퟤ ∞ 피[혨(혟)] = ∫ 혨(혹)혧 (혹)혥혹 • LOTUS with continuous r.v.s: −∞ 혟 • Variance of a continuous r.v.s: ∞ 핍[혟] = 피[(혟 − 피[혟])ퟤ] = ∫ (혹 − 피[혟])ퟤ혥혹 −∞ • Linearity and other properties of 피[] and 핍[] still hold! 11 / 16 Expectation for a continuous r.v. • Expectation of a continuous r.v.: ∞ 피[혟] = ∫ 혹혧혟 (혹)혥혹 −∞ ∞ 피[혟] = ∫ 혹혥혍 (혹) • Unifying notation you may see: −∞ • Expectation of a uniform (0,1): 피[혜] = (혢 + 혣)/ퟤ ∞ 피[혨(혟)] = ∫ 혨(혹)혧 (혹)혥혹 • LOTUS with continuous r.v.s: −∞ 혟 • Variance of a continuous r.v.s: ∞ 핍[혟] = 피[(혟 − 피[혟])ퟤ] = ∫ (혹 − 피[혟])ퟤ혥혹 −∞ • Linearity and other properties of 피[] and 핍[] still hold! • In particular, we still have 핍[혟] = 피[혟 ퟤ] − (피[혟])ퟤ 11 / 16 ퟣ 피[혈] = 피[휋혙ퟤ] = ∫ 휋혳 ퟤ혥혳 ퟢ ퟣ = (휋/ퟥ)혳 ퟥ∣ ퟢ = (휋/ퟥ) ⋅ ퟣퟥ − (휋/ퟥ) ⋅ ퟢퟥ = (휋/ퟥ) • For variance, use 핍[혈] = 피[혈ퟤ] − (피[혈])ퟤ: ퟣ ퟣ 피[혈ퟤ] = 피[휋ퟤ혙ퟦ] = ∫ 휋ퟤ혳 ퟦ혥혳 = (휋ퟤ/ퟧ)혳 ퟧ∣ ퟢ ퟢ = (휋ퟤ/ퟧ) ⋅ ퟣퟧ − (휋ퟤ/ퟧ) ⋅ ퟢퟧ = (휋ퟤ/ퟧ) • ⇝ 핍[혈] = ퟦ휋ퟤ/ퟤퟧ. Challenge: find the c.d.f. and p.d.f. of 혈 • What are 피[혈] and 핍[혈]? • For expectation, use LOTUS! Expectation of random circle areas • Let 혙 ∼ Unif(ퟢ, ퟣ) and 혈 be the area of the circle with radius 혙. 12 / 16 ퟣ 피[혈] = 피[휋혙ퟤ] = ∫ 휋혳 ퟤ혥혳 ퟢ ퟣ = (휋/ퟥ)혳 ퟥ∣ ퟢ = (휋/ퟥ) ⋅ ퟣퟥ − (휋/ퟥ) ⋅ ퟢퟥ = (휋/ퟥ) • For variance, use 핍[혈] = 피[혈ퟤ] − (피[혈])ퟤ: ퟣ ퟣ 피[혈ퟤ] = 피[휋ퟤ혙ퟦ] = ∫ 휋ퟤ혳 ퟦ혥혳 = (휋ퟤ/ퟧ)혳 ퟧ∣ ퟢ ퟢ = (휋ퟤ/ퟧ) ⋅ ퟣퟧ − (휋ퟤ/ퟧ) ⋅ ퟢퟧ = (휋ퟤ/ퟧ) • ⇝ 핍[혈] = ퟦ휋ퟤ/ퟤퟧ. Challenge: find the c.d.f. and p.d.f. of 혈 • For expectation, use LOTUS! Expectation of random circle areas • Let 혙 ∼ Unif(ퟢ, ퟣ) and 혈 be the area of the circle with radius 혙. • What are 피[혈] and 핍[혈]? 12 / 16 • For variance, use 핍[혈] = 피[혈ퟤ] − (피[혈])ퟤ: ퟣ ퟣ 피[혈ퟤ] = 피[휋ퟤ혙ퟦ] = ∫ 휋ퟤ혳 ퟦ혥혳 = (휋ퟤ/ퟧ)혳 ퟧ∣ ퟢ ퟢ = (휋ퟤ/ퟧ) ⋅ ퟣퟧ − (휋ퟤ/ퟧ) ⋅ ퟢퟧ = (휋ퟤ/ퟧ) • ⇝ 핍[혈] = ퟦ휋ퟤ/ퟤퟧ. Challenge: find the c.d.f. and p.d.f. of 혈 ퟣ 피[혈] = 피[휋혙ퟤ] = ∫ 휋혳 ퟤ혥혳 ퟢ ퟣ = (휋/ퟥ)혳 ퟥ∣ ퟢ = (휋/ퟥ) ⋅ ퟣퟥ − (휋/ퟥ) ⋅ ퟢퟥ = (휋/ퟥ) Expectation of random circle areas • Let 혙 ∼ Unif(ퟢ, ퟣ) and 혈 be the area of the circle with radius 혙. • What are 피[혈] and 핍[혈]? • For expectation, use LOTUS! 12 / 16 • For variance, use 핍[혈] = 피[혈ퟤ] − (피[혈])ퟤ: ퟣ ퟣ 피[혈ퟤ] = 피[휋ퟤ혙ퟦ] = ∫ 휋ퟤ혳 ퟦ혥혳 = (휋ퟤ/ퟧ)혳 ퟧ∣ ퟢ ퟢ = (휋ퟤ/ퟧ) ⋅ ퟣퟧ − (휋ퟤ/ퟧ) ⋅ ퟢퟧ = (휋ퟤ/ퟧ) • ⇝ 핍[혈] = ퟦ휋ퟤ/ퟤퟧ. Challenge: find the c.d.f. and p.d.f. of 혈 ퟣ = (휋/ퟥ)혳 ퟥ∣ ퟢ = (휋/ퟥ) ⋅ ퟣퟥ − (휋/ퟥ) ⋅ ퟢퟥ = (휋/ퟥ) Expectation of random circle areas • Let 혙 ∼ Unif(ퟢ, ퟣ) and 혈 be the area of the circle with radius 혙. • What are 피[혈] and 핍[혈]? • For expectation, use LOTUS! ퟣ 피[혈] = 피[휋혙ퟤ] = ∫ 휋혳 ퟤ혥혳 ퟢ 12 / 16 • For variance, use 핍[혈] = 피[혈ퟤ] − (피[혈])ퟤ: ퟣ ퟣ 피[혈ퟤ] = 피[휋ퟤ혙ퟦ] = ∫ 휋ퟤ혳 ퟦ혥혳 = (휋ퟤ/ퟧ)혳 ퟧ∣ ퟢ ퟢ = (휋ퟤ/ퟧ) ⋅ ퟣퟧ − (휋ퟤ/ퟧ) ⋅ ퟢퟧ = (휋ퟤ/ퟧ) • ⇝ 핍[혈] = ퟦ휋ퟤ/ퟤퟧ. Challenge: find the c.d.f. and p.d.f. of 혈 = (휋/ퟥ) ⋅ ퟣퟥ − (휋/ퟥ) ⋅ ퟢퟥ = (휋/ퟥ) Expectation of random circle areas • Let 혙 ∼ Unif(ퟢ, ퟣ) and 혈 be the area of the circle with radius 혙. • What are 피[혈] and 핍[혈]? • For expectation, use LOTUS! ퟣ 피[혈] = 피[휋혙ퟤ] = ∫ 휋혳 ퟤ혥혳 ퟢ ퟣ = (휋/ퟥ)혳 ퟥ∣ ퟢ 12 / 16 • For variance, use 핍[혈] = 피[혈ퟤ] − (피[혈])ퟤ: ퟣ ퟣ 피[혈ퟤ] = 피[휋ퟤ혙ퟦ] = ∫ 휋ퟤ혳 ퟦ혥혳 = (휋ퟤ/ퟧ)혳 ퟧ∣ ퟢ ퟢ = (휋ퟤ/ퟧ) ⋅ ퟣퟧ − (휋ퟤ/ퟧ) ⋅ ퟢퟧ = (휋ퟤ/ퟧ) • ⇝ 핍[혈] = ퟦ휋ퟤ/ퟤퟧ. Challenge: find the c.d.f. and p.d.f. of 혈 Expectation of random circle areas • Let 혙 ∼ Unif(ퟢ, ퟣ) and 혈 be the area of the circle with radius 혙. • What are 피[혈] and 핍[혈]? • For expectation, use LOTUS! ퟣ 피[혈] = 피[휋혙ퟤ] = ∫ 휋혳 ퟤ혥혳 ퟢ ퟣ = (휋/ퟥ)혳 ퟥ∣ ퟢ = (휋/ퟥ) ⋅ ퟣퟥ − (휋/ퟥ) ⋅ ퟢퟥ = (휋/ퟥ) 12 / 16 • ⇝ 핍[혈] = ퟦ휋ퟤ/ퟤퟧ. Challenge: find the c.d.f. and p.d.f. of 혈 ퟣ ퟣ 피[혈ퟤ] = 피[휋ퟤ혙ퟦ] = ∫ 휋ퟤ혳 ퟦ혥혳 = (휋ퟤ/ퟧ)혳 ퟧ∣ ퟢ ퟢ = (휋ퟤ/ퟧ) ⋅ ퟣퟧ − (휋ퟤ/ퟧ) ⋅ ퟢퟧ = (휋ퟤ/ퟧ) Expectation of random circle areas • Let 혙 ∼ Unif(ퟢ, ퟣ) and 혈 be the area of the circle with radius 혙. • What are 피[혈] and 핍[혈]? • For expectation, use LOTUS! ퟣ 피[혈] = 피[휋혙ퟤ] = ∫ 휋혳 ퟤ혥혳 ퟢ ퟣ = (휋/ퟥ)혳 ퟥ∣ ퟢ = (휋/ퟥ) ⋅ ퟣퟥ − (휋/ퟥ) ⋅ ퟢퟥ = (휋/ퟥ) • For variance, use 핍[혈] = 피[혈ퟤ] − (피[혈])ퟤ: 12 / 16 • ⇝ 핍[혈] = ퟦ휋ퟤ/ퟤퟧ. Challenge: find the c.d.f. and p.d.f. of 혈 ퟣ = (휋ퟤ/ퟧ)혳 ퟧ∣ ퟢ = (휋ퟤ/ퟧ) ⋅ ퟣퟧ − (휋ퟤ/ퟧ) ⋅ ퟢퟧ = (휋ퟤ/ퟧ) Expectation of random circle areas • Let 혙 ∼ Unif(ퟢ, ퟣ) and 혈 be the area of the circle with radius 혙. • What are 피[혈] and 핍[혈]? • For expectation, use LOTUS! ퟣ 피[혈] = 피[휋혙ퟤ] = ∫ 휋혳 ퟤ혥혳 ퟢ ퟣ = (휋/ퟥ)혳 ퟥ∣ ퟢ = (휋/ퟥ) ⋅ ퟣퟥ − (휋/ퟥ) ⋅ ퟢퟥ = (휋/ퟥ) • For variance, use 핍[혈] = 피[혈ퟤ] − (피[혈])ퟤ: ퟣ 피[혈ퟤ] = 피[휋ퟤ혙ퟦ] = ∫ 휋ퟤ혳 ퟦ혥혳 ퟢ 12 / 16 • ⇝ 핍[혈] = ퟦ휋ퟤ/ퟤퟧ. Challenge: find the c.d.f. and p.d.f. of 혈 = (휋ퟤ/ퟧ) ⋅ ퟣퟧ − (휋ퟤ/ퟧ) ⋅ ퟢퟧ = (휋ퟤ/ퟧ) Expectation of random circle areas • Let 혙 ∼ Unif(ퟢ, ퟣ) and 혈 be the area of the circle with radius 혙. • What are 피[혈] and 핍[혈]? • For expectation, use LOTUS! ퟣ 피[혈] = 피[휋혙ퟤ] = ∫ 휋혳 ퟤ혥혳 ퟢ ퟣ = (휋/ퟥ)혳 ퟥ∣ ퟢ = (휋/ퟥ) ⋅ ퟣퟥ − (휋/ퟥ) ⋅ ퟢퟥ = (휋/ퟥ) • For variance, use 핍[혈] = 피[혈ퟤ] − (피[혈])ퟤ: ퟣ ퟣ 피[혈ퟤ] = 피[휋ퟤ혙ퟦ] = ∫ 휋ퟤ혳 ퟦ혥혳 = (휋ퟤ/ퟧ)혳 ퟧ∣ ퟢ ퟢ 12 / 16 • ⇝ 핍[혈] = ퟦ휋ퟤ/ퟤퟧ. Challenge: find the c.d.f. and p.d.f. of 혈 Expectation of random circle areas • Let 혙 ∼ Unif(ퟢ, ퟣ) and 혈 be the area of the circle with radius 혙. • What are 피[혈] and 핍[혈]? • For expectation, use LOTUS! ퟣ 피[혈] = 피[휋혙ퟤ] = ∫ 휋혳 ퟤ혥혳 ퟢ ퟣ = (휋/ퟥ)혳 ퟥ∣ ퟢ = (휋/ퟥ) ⋅ ퟣퟥ − (휋/ퟥ) ⋅ ퟢퟥ = (휋/ퟥ) • For variance, use 핍[혈] = 피[혈ퟤ] − (피[혈])ퟤ: ퟣ ퟣ 피[혈ퟤ] = 피[휋ퟤ혙ퟦ] = ∫ 휋ퟤ혳 ퟦ혥혳 = (휋ퟤ/ퟧ)혳 ퟧ∣ ퟢ ퟢ = (휋ퟤ/ퟧ) ⋅ ퟣퟧ − (휋ퟤ/ퟧ) ⋅ ퟢퟧ = (휋ퟤ/ퟧ) 12 / 16 Expectation of random circle areas • Let 혙 ∼ Unif(ퟢ, ퟣ) and 혈 be the area of the circle with radius 혙. • What are 피[혈] and 핍[혈]? • For expectation, use LOTUS! ퟣ 피[혈] = 피[휋혙ퟤ] = ∫ 휋혳 ퟤ혥혳 ퟢ ퟣ = (휋/ퟥ)혳 ퟥ∣ ퟢ = (휋/ퟥ) ⋅ ퟣퟥ − (휋/ퟥ) ⋅ ퟢퟥ = (휋/ퟥ) • For variance, use 핍[혈] = 피[혈ퟤ] − (피[혈])ퟤ: ퟣ ퟣ 피[혈ퟤ] = 피[휋ퟤ혙ퟦ] = ∫ 휋ퟤ혳 ퟦ혥혳 = (휋ퟤ/ퟧ)혳 ퟧ∣ ퟢ ퟢ = (휋ퟤ/ퟧ) ⋅ ퟣퟧ − (휋ퟤ/ퟧ) ⋅ ퟢퟧ = (휋ퟤ/ퟧ) • ⇝ 핍[혈] = ퟦ휋ퟤ/ퟤퟧ. Challenge: find the c.d.f. and p.d.f. of 혈 12 / 16 • 혍 −ퟣ(훼) is the value of 혟 such that ℙ(혟 ≤ 혹) = 훼 • Takes probabilities as arguments! • 혍 −ퟣ(ퟢ.ퟧ) is the median, 혍 −ퟣ(ퟢ.ퟤퟧ) is the lower quartile, etc • Intuition: exactly the same as percentiles on exams. • You’ve probably used them before: confidence interval critical values. Quantile function • Inverse of the c.d.f. 혍 −ퟣ is called the quantile function 13 / 16 • Takes probabilities as arguments! • 혍 −ퟣ(ퟢ.ퟧ) is the median, 혍 −ퟣ(ퟢ.ퟤퟧ) is the lower quartile, etc • Intuition: exactly the same as percentiles on exams. • You’ve probably used them before: confidence interval critical values. Quantile function • Inverse of the c.d.f. 혍 −ퟣ is called the quantile function • 혍 −ퟣ(훼) is the value of 혟 such that ℙ(혟 ≤ 혹) = 훼 13 / 16 • 혍 −ퟣ(ퟢ.ퟧ) is the median, 혍 −ퟣ(ퟢ.ퟤퟧ) is the lower quartile, etc • Intuition: exactly the same as percentiles on exams. • You’ve probably used them before: confidence interval critical values. Quantile function • Inverse of the c.d.f. 혍 −ퟣ is called the quantile function • 혍 −ퟣ(훼) is the value of 혟 such that ℙ(혟 ≤ 혹) = 훼 • Takes probabilities as arguments! 13 / 16 • Intuition: exactly the same as percentiles on exams. • You’ve probably used them before: confidence interval critical values. Quantile function • Inverse of the c.d.f. 혍 −ퟣ is called the quantile function • 혍 −ퟣ(훼) is the value of 혟 such that ℙ(혟 ≤ 혹) = 훼 • Takes probabilities as arguments! • 혍 −ퟣ(ퟢ.ퟧ) is the median, 혍 −ퟣ(ퟢ.ퟤퟧ) is the lower quartile, etc 13 / 16 • You’ve probably used them before: confidence interval critical values. Quantile function • Inverse of the c.d.f. 혍 −ퟣ is called the quantile function • 혍 −ퟣ(훼) is the value of 혟 such that ℙ(혟 ≤ 혹) = 훼 • Takes probabilities as arguments! • 혍 −ퟣ(ퟢ.ퟧ) is the median, 혍 −ퟣ(ퟢ.ퟤퟧ) is the lower quartile, etc • Intuition: exactly the same as percentiles on exams. 13 / 16 Quantile function • Inverse of the c.d.f. 혍 −ퟣ is called the quantile function • 혍 −ퟣ(훼) is the value of 혟 such that ℙ(혟 ≤ 혹) = 훼 • Takes probabilities as arguments! • 혍 −ퟣ(ퟢ.ퟧ) is the median, 혍 −ퟣ(ퟢ.ퟤퟧ) is the lower quartile, etc • Intuition: exactly the same as percentiles on exams. • You’ve probably used them before: confidence interval critical values. 13 / 16 Quantile functions 1.0 0.8 0.6 F(x) 0.4 0.2 0.0 -4 -2 0 2 4 x 14 / 16 • Not 혍 (혟) ≠ ℙ(혟 ≤ 혟). 1. Let 혜 ∼ Unif(ퟢ, ퟣ) and 혟 = 혍 −ퟣ(혜), then 혟 is an r.v. with c.d.f. 혍 . 2. If 혟 is an r.v. with c.d.f. 혍 , then 혍 (혟) ∼ Unif(ퟢ, ퟣ). • Careful: 혍 (혟) means plug the random variable into the c.d.f. as a function. Universality of the Uniform • The Uniform distribution has a deep connection to all continuous r.v.s 15 / 16 • Not 혍 (혟) ≠ ℙ(혟 ≤ 혟). 2. If 혟 is an r.v. with c.d.f. 혍 , then 혍 (혟) ∼ Unif(ퟢ, ퟣ). • Careful: 혍 (혟) means plug the random variable into the c.d.f. as a function. Universality of the Uniform • The Uniform distribution has a deep connection to all continuous r.v.s 1. Let 혜 ∼ Unif(ퟢ, ퟣ) and 혟 = 혍 −ퟣ(혜), then 혟 is an r.v. with c.d.f. 혍 . 15 / 16 • Not 혍 (혟) ≠ ℙ(혟 ≤ 혟). • Careful: 혍 (혟) means plug the random variable into the c.d.f. as a function. Universality of the Uniform • The Uniform distribution has a deep connection to all continuous r.v.s 1. Let 혜 ∼ Unif(ퟢ, ퟣ) and 혟 = 혍 −ퟣ(혜), then 혟 is an r.v. with c.d.f. 혍 . 2. If 혟 is an r.v. with c.d.f. 혍 , then 혍 (혟) ∼ Unif(ퟢ, ퟣ). 15 / 16 • Not 혍 (혟) ≠ ℙ(혟 ≤ 혟). Universality of the Uniform • The Uniform distribution has a deep connection to all continuous r.v.s 1. Let 혜 ∼ Unif(ퟢ, ퟣ) and 혟 = 혍 −ퟣ(혜), then 혟 is an r.v. with c.d.f. 혍 . 2. If 혟 is an r.v. with c.d.f. 혍 , then 혍 (혟) ∼ Unif(ퟢ, ퟣ). • Careful: 혍 (혟) means plug the random variable into the c.d.f. as a function. 15 / 16 Universality of the Uniform • The Uniform distribution has a deep connection to all continuous r.v.s 1. Let 혜 ∼ Unif(ퟢ, ퟣ) and 혟 = 혍 −ퟣ(혜), then 혟 is an r.v. with c.d.f. 혍 . 2. If 혟 is an r.v. with c.d.f. 혍 , then 혍 (혟) ∼ Unif(ퟢ, ퟣ). • Careful: 혍 (혟) means plug the random variable into the c.d.f. as a function. • Not 혍 (혟) ≠ ℙ(혟 ≤ 혟). 15 / 16 • Doesn’t necessarily hold for discrete r.v.s Symmetry of iid continuous r.v.s Proposition Let 혟ퟣ, … , 혟혯 be i.i.d. from a continuous distribution. Then, ퟣ ℙ(혟 < 혟 < ⋯ < 혟 ) = 혢ퟣ 혢ퟤ 혢혯 혯! for any permutation 혢ퟣ, 혢ퟤ, … , 혢혯 of ퟣ, ퟤ, … , 혯. • All orderings of continuous i.i.d. r.v.s are equally likely. 16 / 16 Symmetry of iid continuous r.v.s Proposition Let 혟ퟣ, … , 혟혯 be i.i.d. from a continuous distribution. Then, ퟣ ℙ(혟 < 혟 < ⋯ < 혟 ) = 혢ퟣ 혢ퟤ 혢혯 혯! for any permutation 혢ퟣ, 혢ퟤ, … , 혢혯 of ퟣ, ퟤ, … , 혯. • All orderings of continuous i.i.d. r.v.s are equally likely. • Doesn’t necessarily hold for discrete r.v.s 16 / 16