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Consider independent trials, each of which results in outcome i, i = 0, 1, ... , k, with probability pi, k i=0 pi = 1. Let N denote the number of trials needed to obtain an outcome that is not equal to 0, and let X be that outcome.

(a) Find P{N = n}, n Ú 1.

(b) Find P{X = j}, j = 1, ... , k.

(c) Show that P{N = n, X = j} = P{N = n}P{X = j}.

(d) Is it intuitive to you that N is independent of X?

(e) Is it intuitive to you that X is independent of N?

Short Answer

Expert verified

a. P(N=n)=p0n-1.(1-p0)

b. P[X=j]=pj1-p0

c. P(X=j,N=n)=P(X=j).P(N=n),where X and N are independent.

d. Yes, N is independent of X.

e. Yes, X is independent of N.

Step by step solution

01

Content Introduction

Observe that random variables X and N are independent.

02

Explanation (Part a)

We observe that N has geometric distribution with parameter of success 1 - p0 .

Hence, P(N=n)=p0n-1.(1-p0).

This is because of the fact that N = n, it implies that in first n - 1 trials we have obtained outcomes equal to zero and in the nth trial we have obtained any other trial rather than zero.

03

Stepp 3: Explanation (Part b)

Whatever is the time of first non zero outcome, every other non-zero outcome has proportional probability to be that outcome,

P(X=j)=pj1-p0

Observe that we divide pj with 1 - p0since that we know considered outcome cannot be zero and we want sum of all probabilities is equal to one.

04

Explanation (part c)

We have that,

P(X=j,N=n)=P(X=j/N=n).P(N=n)

Since the probability P(X=j)is equal no matter which time of first non zero outcome is, we have that P(X=j/N=n)=P(X=j)

Hence, P(X=j,N=n)=P(X=j).P(N=n).

So, we have proved that X and N are independent.

05

Explanation (Part d)

Yes, as its known what is the outcome of first non-zero outcome does not imply when the first non zero outcome might happen. Hence, N is independent of X.

06

Explanation (Part e)

Yes, as its know the time the first non zero outcome does not tell us what is that outcome. The probabilities remain the same. Hence, X is independent of N.

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Most popular questions from this chapter

The joint probability density function of X and Y is given by

f(x.y)=67(x2+xy2)0<x<1,0<y<2

(a) Verify that this is indeed a joint density function.

(b) Compute the density function of X.

(c) Find P{X > Y}.

(d) Find P{Y > 1 2 |X < 1 2 }.

(e) Find E[X].

(f) Find E[Y].

The accompanying dartboard is a square whose sides are of length 6.

The three circles are all centered at the center of the board and are of radii 1, 2, and 3, respectively. Darts landing within the circle of radius 1 score 30 points, those landing outside this circle, but within the circle of radius 2, are worth 20 points, and those landing outside the circle of radius 2, but within the circle of radius 3, are worth 10 points. Darts that do not land within the circle of radius 3 do not score any points. Assuming that each dart that you throw will, independently of what occurred on your previous throws, land on a point uniformly distributed in the square, find the probabilities of the accompanying events:

(a) You score 20 on a throw of the dart.

(b) You score at least 20 on a throw of the dart.

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Consider a sequence of independent Bernoulli trials, each of which is a success with probability p. Let X1 be the number of failures preceding the first success, and let X2 be the number of failures between the first two successes. Find the joint mass function of X1 and X2.

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(a) Y2=1

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