Here,
The probability mass function of N is
Now,
4.13 Let N be a geometric random variable with parameter p - 1/3. Calculate Pr[N s...
[Probability] Let N be a geometric random variable with parameter p. Given N,generate N many i.i.d. random numbers U1, U2, . . . , UN uniformly from [0,1]. Let M= max 1≤i≤N Ui. Find the cdf of M, i.e., find P(M≤x).
3. Let X be a random variable from a geometric distribution with parameter p (P(X- k p(1-P)"-, } k-1 k-1, 2, ...). Find Emin{X, 100
Let X be the random variable with the geometric distribution with parameter 0 < p < 1. (1) For any integer n ≥ 0, find P(X > n). (2) Show that for any integers m ≥ 0 and n ≥ 0, P(X > n + m|X > m) = P(X > n) (This is called memoryless property since this conditional probability does not depend on m.)
A discrete random variable X follows the geometric distribution with parameter p, written X ∼ Geom(p), if its distribution function is A discrete random variable X follows the geometric distribution with parameter p, written X Geom(p), if its distribution function is 1x(z) = p(1-P)"-1, ze(1, 2, 3, ). The Geometric distribution is used to model the number of flips needed before a coin with probability p of showing Heads actually shows Heads. a) Show that fx(x) is indeed a probability...
4. (9 pts) Suppose the random variable Y has a geometric distribution with parameter p. Let ?? = √?? 3 3 . Find the probability distribution of V 3 4. (9 pts) Suppose the random variable Y has a geometric distribution with parameter p. Let V 3 Find the probability distribution of.
Question 1: 1a) Let the random variable X have a geometric distribution with parameter p , i.e., P(X = x) = pq??, x=1,2,... i) Show that P(X > m)=q" , where m is a positive integer. (5 points) ii) Show that P(X > m+n X > m) = P(X>n), where m and n are positive integers. (5 points) 1b) Suppose the random variable X takes non-negative integer values, i.e., X is a count random variable. Prove that (6 points) E(X)=...
3. Let X be a geometric random variable with parameter p. Prove that P(X >k+r|X > k) = P(X > r). This is called the memoryless property of the geometric random variable.
Problem 8 (10 points). Let X be the random variable with the geometric distribution with parameter 0 <p <1. (1) For any integer n > 0, find P(X >n). (2) Show that for any integers m > 0 and n > 0, P(X n + m X > m) = P(X>n) (This is called memoryless property since this conditional probability does not depend on m. Dobs inta T obabilita ndomlu abonn liaht bulb indofootin W
2. (Ross 3.2) Let Xi and X2 be independent geometric random variables having the same parameter p. (a) Compute the pmf for the random variable Y (b) Compute Pr(X,-iX, +X2=n) - Xi+ X2
3. Consider a discrete random variable X which follows the geometric distribution f(x,p) = pr-1(1-p), x = 1.2. . . . , 0 < p < 1. Recall that E(x) (1-p) (a) Find the Fisher information I(p). (b) Show that the Cramer-Rao inequality is strict e) Let XX ~X. Find the maximum likelihood estimator of p. Note that the expression you find may look complicated and hard to evaluate. (d) Now modify your view by setting μ T1p such that...