Solution Found!
Let Y1 have an exponential distribution with mean ? and
Chapter 5, Problem 141E(choose chapter or problem)
Let \(Y_{1}\) have an exponential distribution with mean \(\lambda\) and the conditional density of \(Y_{2}\) given \(Y_{1}=y_{1}\) be
\(f\left(y_{2} \mid y_{1}\right)=\left\{\begin{array}{ll}
1 / y_{1}, & 0 \leq y_{2} \leq y_{1} \\
0, & \text { elsewhere. }
\end{array}\right.\)
Find \(E\left(Y_{2}\right)\) and \(V\left(Y_{2}\right)\), the unconditional mean and variance of \(Y_{2}\).
Equation Transcription:
Text Transcription:
Y_1
\lambda
Y_2
Y_1=y_1
f(y_2 | y_1)={1/y1 0 \leq y_2 \leq y_1 0 elsewhere
E(Y_2)
V(Y_2)
Y_2
Questions & Answers
QUESTION:
Let \(Y_{1}\) have an exponential distribution with mean \(\lambda\) and the conditional density of \(Y_{2}\) given \(Y_{1}=y_{1}\) be
\(f\left(y_{2} \mid y_{1}\right)=\left\{\begin{array}{ll}
1 / y_{1}, & 0 \leq y_{2} \leq y_{1} \\
0, & \text { elsewhere. }
\end{array}\right.\)
Find \(E\left(Y_{2}\right)\) and \(V\left(Y_{2}\right)\), the unconditional mean and variance of \(Y_{2}\).
Equation Transcription:
Text Transcription:
Y_1
\lambda
Y_2
Y_1=y_1
f(y_2 | y_1)={1/y1 0 \leq y_2 \leq y_1 0 elsewhere
E(Y_2)
V(Y_2)
Y_2
ANSWER:
Solution :
Step 1 of 1:
Let denotes an exponential distribution with mean
.
Then is
Our goal is:
We need to find E() and V(
), the unconditional mean and variance of
.
Now we have to find E() and V(
), the unconditional mean and variance of
.