Question:
Let (X, Y) be a random vector having the joint probability density function
\(f(x,y)=\begin{cases} \frac{\sqrt2}{\sqrt{\pi}}e^{-2x}e^{-\frac{(y-x)^2}{2}}, & 0 <x< ∞,-∞<y<∞\\ 0,& \text{otherwise} \end{cases}\)
Then E(Y) equals
Let (X, Y) be a random vector having the joint probability density function
\(f(x,y)=\begin{cases} \frac{\sqrt2}{\sqrt{\pi}}e^{-2x}e^{-\frac{(y-x)^2}{2}}, & 0 <x< ∞,-∞<y<∞\\ 0,& \text{otherwise} \end{cases}\)
Then E(Y) equals
\(f(x,y)=\begin{cases} \frac{\sqrt2}{\sqrt{\pi}}e^{-2x}e^{-\frac{(y-x)^2}{2}}, & 0 <x< ∞,-∞<y<∞\\ 0,& \text{otherwise} \end{cases}\)
Then E(Y) equals
Updated On: Oct 1, 2024
- \(\frac{1}{2}\)
- 2
- 1
- \(\frac{1}{4}\)
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The Correct Option is A
Solution and Explanation
The correct option is (A) : \(\frac{1}{2}\).
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