Q2(iv.) If X1, X2, · are independent and identically distributed exponential random variables with parameter A and N is a geometric random variable with parameter p independent of the sequence X1, X2, ... (ie P(N =n) = p(1 – p)"-1), then show that N Y =X; i=1 s exponentially distributed with parameter Xp using the moment generating function from part (i). Hint: The m.g.f. of an exponential random variable with parameter A is A/(A – t).
Q2(iv.) If X1, X2, · are independent and identically distributed exponential random variables with parameter A and N is a geometric random variable with parameter p independent of the sequence X1, X2, ... (ie P(N =n) = p(1 – p)"-1), then show that N Y =X; i=1 s exponentially distributed with parameter Xp using the moment generating function from part (i). Hint: The m.g.f. of an exponential random variable with parameter A is A/(A – t).
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.8: Probability
Problem 19E
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Question
Could you solve (iv)?
Thank you.
![Q2 Let X1, X2, -- be independent and identically distributed random variables. Let N be a non-negative, integer
valued random variable that is independent of the sequence X;, i > 1. Let
...
N
Y =X;.
Q2(i.) Let øy (t) = E[eY] be the moment generating function of the random variable Y. Using conditioning
argument (i.e., E[X] = E[E[X[Y]) show that
ør (t) = E[(øx(t))^],
where øx (t) is the m.g.f. of X.
Q2 (ii.) Using the moment generating function from part (i) show that
E[Y] = E[N]E[X]
Q2 (iii.) Using the moment generating function from part (1) and the results from part (i) show that
Var[Y]
= E[N]Var[X] + (E[X])²Var[N].
Q2(iv.) If X1, X2, ·. are independent and identically distributed exponential random variables with parameter A
and N is a geometric random variable with parameter p independent of the sequence X1, X2, ... (i.e
P(N = n) = p(1 – p)"-1), then show that
N
Y =X;
is exponentially distributed with parameter Xp using the moment generating function from part (i).
Hint: The m.g.f. of an exponential random variable with parameter A is A/(A – t).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf4d5614-e5fa-4399-aabc-c345eeef0588%2F6173f788-5957-4d89-8061-03c539ce0de2%2Ftttq8y_processed.png&w=3840&q=75)
Transcribed Image Text:Q2 Let X1, X2, -- be independent and identically distributed random variables. Let N be a non-negative, integer
valued random variable that is independent of the sequence X;, i > 1. Let
...
N
Y =X;.
Q2(i.) Let øy (t) = E[eY] be the moment generating function of the random variable Y. Using conditioning
argument (i.e., E[X] = E[E[X[Y]) show that
ør (t) = E[(øx(t))^],
where øx (t) is the m.g.f. of X.
Q2 (ii.) Using the moment generating function from part (i) show that
E[Y] = E[N]E[X]
Q2 (iii.) Using the moment generating function from part (1) and the results from part (i) show that
Var[Y]
= E[N]Var[X] + (E[X])²Var[N].
Q2(iv.) If X1, X2, ·. are independent and identically distributed exponential random variables with parameter A
and N is a geometric random variable with parameter p independent of the sequence X1, X2, ... (i.e
P(N = n) = p(1 – p)"-1), then show that
N
Y =X;
is exponentially distributed with parameter Xp using the moment generating function from part (i).
Hint: The m.g.f. of an exponential random variable with parameter A is A/(A – t).
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