$Foundations in probability$

Question

$Foundations in probability$

Check picture please

Probability

Burgoos

226

Matchmaticians Foundations in probability File #1

File #1 (PNG)

Report

Burgoos

0

philip can you check my other question as well, its similar but my deadline is wrong
Erdos

0

I will.

Answer

The answer is accepted.

Join Matchmaticians Affiliate Marketing Program to earn up to 50% commission on every question your affiliated users ask or answer.

Burgoos

0

philip can you check my other question as well, its similar but my deadline is wrong

Answer 1

Answers can be viewed only if

The questioner was satisfied and accepted the answer, or
The answer was disputed, but the judge evaluated it as 100% correct.

View the answer

We have
\[f_Y(y)=\int_{0}^{\infty}\frac{1}{2}ye^{-xy}dx=-\frac{1}{2}e^{-xy}|_{x=0}^{x=\infty}=0+\frac{1}{2}=\frac{1}{2}.\]
Then
\[f(x|y)=\frac{f(x,y)}{f_Y(y)}=\frac{\frac{1}{2}ye^{-xy}}{\frac{1}{2}}=ye^{-xy}.\]
So
\[E[e^{X/2}|Y=1]=\int_0^{\infty}e^{x/2}f(x|y=1)dx=\int_0^{\infty}e^{x/2}(1e^{-x\cdot 1})dx=\int_0^{\infty}e^{-x/2}dx\]
\[=-2e^{-x/2}|_0^{\infty}=0-(-2)=2.\]

Erdos

4.4K

$Foundations in probability$

Answer

Related Questions

Search