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Lecture 6: Special Probability Distributions

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Lecture 6: Special Probability Distributions The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di Lecture 6: Special Probability Distributions Assist. Prof. Dr. Emel YAVUZ DUMAN MCB1007 Introduction to Probability and Statistics Istanbul˙ K¨ult¨ur University The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di Outline 1 The Discrete Uniform Distribution 2 The Bernoulli Distribution 3 The Binomial Distribution 4 The Negative Binomial and Geometric Distribution The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di Outline 1 The Discrete Uniform Distribution 2 The Bernoulli Distribution 3 The Binomial Distribution 4 The Negative Binomial and Geometric Distribution The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di The Discrete Uniform Distribution If a random variable can take on k different values with equal probability, we say that it has a discrete uniform distribution; symbolically, Definition 1 A random variable X has a discrete uniform distribution and it is referred to as a discrete uniform random variable if and only if its probability distribution is given by 1 f (x)= for x = x , x , ··· , xk k 1 2 where xi = xj when i = j. The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di The mean and the variance of this distribution are Mean: k k 1 x1 + x2 + ···+ xk μ = E[X ]= xi f (xi ) = xi · = , k k i=1 i=1 1/k Variance: k 2 2 2 1 σ = E[(X − μ) ]= (xi − μ) · k i=1 2 2 2 (x − μ) +(x − μ) + ···+(xk − μ) = 1 2 k The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di In the special case where xi = i, the discrete uniform distribution 1 becomes f (x)= k for x =1, 2, ··· , k, and in this from it applies, for example, to the number of points we roll with a balanced die. Example 2 If X has the discrete uniform distribution f (x)=1/k for x =1, 2, ··· , k, show that μ k+1 (a) its mean is = 2 ; σ2 k2−1 (b) its variance is = 12 . The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di Solution. (a) Mean: k k 1 1+2+···+ k μ = E(X)= if (x)= i · = k k i=1 i=1 k k ( +1) k +1 = 2 = . k 2 σ2 μ − μ2 2 − 2 (b) Variance: = 2 = E(X ) [E(X)] k k 1 12 +22 + ···+ k2 μ = E(X 2)= i 2f (x)= i 2 · = 2 k k i=1 i=1 k k k ( +1)(2 +1) (k + 1)(2k +1) = 6 = , k 6 (k + 1)(2k +1) k +1 2 k2 − 1 σ2 = − = . 6 2 12 The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di Example 3 If X has the discrete uniform distribution f (x)=1/k for x =1, 2, ··· , k, show that its moment-generating function is given by et (1 − ekt ) MX (t)= . k(1 − et ) Also find the mean of this distribution by evaluating limt→0 MX (t), compare the results with that obtained in Example 2. The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di Solution. We know that k t 2t kt tX it 1 e + e + ···+ e MX (t)=E(e )= e = . k k i=1 On the other hand, since S = et + e2t + e3t + ···+ e(k−1)t + ekt −etS = −e2t − e3t − e4t − ··· −ekt − e(k+1)t ⇒ S(1 − et )=S − et S = et − e(k+1)t = et(1 − ekt ) ⇒ S(1 − et ) et (1 − ekt ) = S = et + e2t + e3t + ···+ e(k−1)t + ekt = 1 − et 1 − et thus et + e2t + ···+ ekt et (1 − ekt ) MX (t)= = . k k(1 − et ) The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di et + e2t + ···+ ekt MX (t)= ⇒ k et +2e2t + ···+ ke kt M (t)= ⇒ X k ··· k(k+1) 1+2+ + k 2 k +1 μ = lim MX (t)= = = . t→0 k k 2 The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di Outline 1 The Discrete Uniform Distribution 2 The Bernoulli Distribution 3 The Binomial Distribution 4 The Negative Binomial and Geometric Distribution The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di The Bernoulli Distribution If an experiment has two possible outcomes, “success” and “failure” and their probabilities are, respectively, θ and 1 − θ,then the number of successes, 0 or 1, has a Bernoulli distribution; symbolically, Definition 4 A random variable X has a Bernoulli distribution and it is referred to as a Bernoulli random variable if and only if its probability distribution is given by f (x; θ)=θx (1 − θ)1−x for x =0, 1. Thus, f (0; θ)=1− θ and f (1; θ)=θ are combined into a single formula. Since the Bernoulli distribution is a special case of the Binomial distribution, we shall not discuss it here in any detail. The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di In connection with the Bernoulli distribution, a success may be getting heads with a balanced coin, it may be catching pneumonia, it may be passing (or failing) an examination, and it may be losing a race. We refer to an experiment to which the Bernoulli distribution applies as a Bernoulli trial,orsimplyatrial,andto sequences of such experiments as repeated trials. Examples. Toss a coin: S = {H, T }. Throw a fair die: S = {face value is a six, face value is not a six }. Sent a message through a network and record whether or not it is received: S = {successful transmission, unsuccessful transmission}. Draw a part from an assembly line and record whether or not it is defective: S = {defective, good}. The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di Example 5 If X has the Bernoulli distribution, show that (a) its mean is μ = θ; (b) its variance is σ2 = θ(1 − θ). Solution. (a) 1 1 μ = E(X)= xf (x; θ)= xθx (1 − θ)1−x x=0 x=0 =0· θ0(1 − θ)1−0 +1· θ1(1 − θ)1−1 = θ (b) 1 1 μ 2 2 θ 2θx − θ 1−x 2 = E(X )= x f (x; )= x (1 ) x=0 x=0 =02 · θ0(1 − θ)1−0 +12 · θ1(1 − θ)1−1 = θ ⇒ σ2 2 − 2 μ − μ2 θ − θ2 θ − θ = E(X ) [E(X)] = 2 = = (1 ) The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di Example 6 Show that for the Bernoulli distribution μr = θ for r =1, 2, 3, ···. Solution. 1 1 tX tx tx x 1−x MX (t)=E(e )= e f (x; θ)= e θ (1 − θ) x=0 x=0 = et·0θ0(1 − θ)1−0 + et·1θ1(1 − θ)1−1 =1− θ + θ · et t2 t3 tr =1+θ(et − 1) = 1 + θ 1+t + + + ···+ + ···−1 2! 3! r! 2 3 r t t t =1+θt + θ + θ + ···+ θ + ···⇒μr = θ. 2! 3! r! The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di Outline 1 The Discrete Uniform Distribution 2 The Bernoulli Distribution 3 The Binomial Distribution 4 The Negative Binomial and Geometric Distribution The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di The Binomial Distribution Repeated trials play a very important role in the probability and statistics, especially when the number of trial is fixed, the parameter θ (the probability of a success) is the same for each trial, and the trials are all independent. The theory that we shall discuss in this section has many applications; for instance, it applies if we want to know the probability of getting 5 heads in 12 flips of a coin, the probability that 7 of 10 persons will recover from a tropical disease, or the probability that 35 of 80 persons will respond to a mail-order solicitation. However, this is the case only if each of the 10 persons has the same chance of recovering from the disease and and their recoveries are independent. The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di To derive a formula for the probability of getting “x successes in n trials” under the stated conditions, observe that the probability of getting x successes and n − x failures in a specific order is θx (1 − θ)n−x . There is one factor θ for each success, one factor 1 − θ for each failure , and the x factors θ and n − x factors 1 − θ are all multiplied together by virtue of the assumption of independence. Since this probability applies to any sequence of n trials in which there are x successes and n − x failures, we have only to count how many sequences of this kind there are and then multiply θx (1 − θ)n−x by that number. Clearly, number of ways in which we can select the x trials on which there is to be a success is n x , and it follows that the desired probability for “x successes in n n x n−x trials” is x θ (1 − θ) .
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