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SOLUTION for HOMEWORK 1, STAT 3372 Welcome to Your First SOLUTION FOR HOMEWORK 1, STAT 3372 Welcome to your first homework. Remember that you are always allowed to use Tables allowed on SOA/CAS exam 4. You can find them on my webpage. Another remark: 6 minutes per problem is your “speed”. Now you probably will not be able to solve your problems so fast — but this is the goal. Try to find mistakes (and get extra points) in my solutions. Typically they are silly arithmetic mistakes (not methodological ones). They allow me to check that you did your HW on your own. Please do not e-mail me about your findings — just mention them on the first page of your solution and count extra points. You can use them to compensate for wrongly solved problems in this homework. They cannot be counted beyond 10 maximum points for each homework. Now let us look at your problems. General Remark: It is always prudent to begin with writing down what is given and what you need to establish. This step can help you to understand/guess a possible solution. Also, your solution should be written neatly so, if you have extra time, it is easier to check your solution. 1. Problem 2.4 Given: The hazard function is h(x)=(A + e2x)I(x ≥ 0) and S(0.4) = 0.5. Find: A Solution: I need to relate h(x) and S(x) to find A. There is a formula on page 17, also discussed in class, which is helpful here. Namely, x − h(u)du S(x)= e 0 . R It allows us to find A. Indeed, x x h(u)du = (A + e2u)du = Ax + (1/2)(e2x − 1). Z0 Z0 Using the given relation S(0.4)=0.5 we get 2x ln(0.5) = ln(S(0.4)) = −[Ax + (1/2)(e − 1)|x=0.4. Now I simplify the last relation and get ln(2) = A(0.4)+(1/2)(e0.8 − 1) and then ln(2) − e0.8/2+1/2 0.70 − 1.11+0.5 A = = =0.20. 0.4 0.4 2. Problem 3.3 Given: σ Var(X) E(X − µ)2 µ := E(X)=2, η := = = = 2; µ′ := E(X3) = 136. µ q µ q µ 3 1 Find the skewness γ1. Solution: First, let us remember that the skewness is µ E(X − µ)3 γ := 3 = . (1) 1 σ3 σ3 The standard deviation σ I can calculate from the coefficient of variation η and the mean µ, namely σ = µη =4. A formula for the third central moment via raw moments is (check it via (a + b)3 = a3 +3a2b +3ab2 + b3, and remember that this is a particular case of a general binomial k k r k−r formula (a + b) = r=0[k!/(r!(k − r)!)]a b ) P E(X − µ)3 = E(X3) − 3E(X2)µ +2µ3. (2) ′ 2 In the right-side of (2) I do not know µ2 := E(X ) but I do know how to calculate it via central moments: E(X2) = Var(X) + [E(X)]2 = σ2 + µ2 =16+4=20. Using this in (2) yields 3 µ3 = E(X − µ) = 136 − (3)(20)(2) + (2)(8) = 32. Then, using (1) we get the wished 3 γ1 = 32/4 =1/2. 3. Problem 3.6 Given that the mean excess loss functions eX (d) and eY (d) are related as eY (30) = eX (30)+4 (3) where X ∼ Unif([0, 100]) and Y ∼ Unif([0,w]), find w. Solution: Let us remember formulae for the mean excess loss function that may be useful here: ∞ ∞ d (z − d)fZ(z)dz d SZ (z)dz eZ (d) := = . (4) R SZ (d) R SZ (d) Above I wrote two possible expressions because one of them can be more helpful (faster to use). Here the second one looks more attractive to me because I only need to know the survival function SZ for a uniform RV Unif([0,u]). Let us calculate it (but if you can find it in the Table - use it!) ∞ −1 −1 SZ (z) = Pr(Z > z)= u I(0 <x<u)dx =(u − z)u I(z ∈ [0,u]) + I(z < 0). Zz This together with (4) yields u −1 2 2 d (u − z)u dz u(u − d) − (1/2)(u − d ) eZ (d)= −1 = R (u − d)u u − d 2 = u − (1/2)(u + d)=(1/2)(u − d). Using this expression to calculate the two excess loss functions in (3) yields (note that here d = 30) eX (30) = (1/2)[100 − 30] = 35, eY (30) = (1/2)[w − 30] = w/2 − 15. Plug in (3) and get 35 = w/2 − 15+4. (5) We get w = 92. Remark: Now is a time to check correctness of the answer. Can w be smaller 100? Does this look right to you? Note that CAS/SOA exams are multiple choice exams, so you understanding of a topic can help and drastically reduce time to solve a problem. Here it is clear that w must be larger than 100 because eY (30) > eX (30)! My mistake was in (5) where I incorrectly plugged-in numbers. A correct step is: w/2 − 15 = 35 + 4 which gives me w = 108. −1 −x/λ 4. Problem 3.7. Given: fX (x) = λ e I(x > 0). [This is exponential RV with the mean λ.] Find the mean excess loss function eX (d) at d = λ. Solution: Using (4) we get ∞ d SX (u)du eX (d)= . R SX (d) For the exponential RV the survival function is for x ≥ 0 ∞ ∞ −1 −u/λ −u/λ ∞ −x/λ S(x)= fX (u)du = λ e du = −e |x = e . Zx Zx Please check that SX has properties of the survival function. Then ∞ −u/λ −u/λ ∞ −d/λ d e du −λe |d λe eX (d)= −d/λ = −d/λ = −d/λ = λ. R e e e What we see is the famous memory-less property of Exponential RV. 5. Problem 3.10. (a) Wrong. Empirical distribution function is discontinuous (it cor- responds to a discrete random variable) and then the mean excess loss function is also discontinuous. (b). Correct, proved earlier. (c). Wrong. Using Table A, p.671, α > 0, the survival function is SX (u) = [θ/(θ + u)]αI(u> 0), α> 0. Thus the mean excess loss function is ∞ α ∞ −α ∞ d SX (u)du θ d (u + θ) du α −α eX (d)= = α =(d + θ) (u + θ) du. R SX (d) R[θ/(d + θ)] Zd Remark: Another way to quickly check a possible answer is to use the formula E(X) − E(X ∧ d) eX (d)= SX (d) 3 and then use the Table. Note that the integral converges only if α> 1, and then α −1 −α+1 eX (d)=(d + θ) (α − 1) (d + θ) =(d + θ)/(α − 1), (6) so it is always increasing in d. 6. Problem 3.11. Remember that in Problem 3.10 I explained that the mean excess loss function (and the mean) for Pareto exists only if α> 1. −3.5 7. Problem 3.13. Given fX (x)=2.5x I(x ≥ 1). Find the coefficient of variation η. Solution: By its definition [Var(X)]1/2 [E(X2) − µ2]1/2 η = = . (7) E(X) µ ∞ −3.5 2 ∞ 2 −3.5 Now we calculate: E(X)= 1 (2.5)xx dx =5/3 and E(X )= 1 (2.5)x x dx =5. Plug in (7) and get R R η = [5 − 25/9]1/2/[5/3] = .9. Remark: you may notice that X is a single-parameter Pareto (see the Table A.4.1.4) with α = −2.5 and fixed (set in advance) θ = 1. Note that for this Pareto the support is x > θ. Do not be confused by another two-parameters Pareto (Pareto part II) described in A.2.3. Here both α and θ are parameters and the difference is that the support is x> 0! So be cautious with Pareto as well as with other distributions — accurately try to figure out which one is related to your problem. By the way, do you think that Y = X − θ where X is the single-parameter and Y is two-parameter Pareto? In any case, because the Table is given, you can use it to save some time. 8. Problem 3.16. Given: Empirical cdf Fˆ(x) is equal to 0.2 at x = 400, 0.7 at x = 800, and 0.1 at x = 1600. Find a corresponding empirical skewness. 3 3 Solution: Remember that skewness is γ1 = E(X − µ) /σ . For empirical one we use the empirical cdf. We need to calculate 3 moments. Let us do this. We do this via corresponding empirical probability mass functionp ˆ(x) which is equal to jumps of the empirical cdf, that is,p ˆ(x) is equal to 0.2, 0.7, and 0.1 at x equal to 400, 800 and 1,600. Now we calculate: Eˆ(X)=ˆµ = xpˆ(x)=(.2)(400) + (.7)(800) + (.1)(1600) = 80+560+160 = 800. x X Further, Eˆ(X − µˆ)2 =(.2)(400 − 800)2 +(.7)(800 − 800)2 +(.1)(1600 − 800)2 = 96, 000. Further, Eˆ(X − µˆ)3 =(.2)(400 − 800)3 +(.7)(800 − 800)3 +(.1)(1600 − 800)3 = 38, 400, 000. Plug in the numbers and get 3/2 γ1 = 38, 400, 000/(96, 000) =1.29. 4 9. Problem 3.17. Given: cdf is F (x)=(1−x−2)I(x ≥ 1).
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