The joint probability density function of X and Y is given by
$\displaystyle
f(x,y) = \begin{cases} exp[-(x+y)] \qquad 0 \le x < \infty \; , \; 0
\le y < \infty \\
0 \qquad , otherwise
\end{cases}
$
Compute E[X] and P[Y>2]
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The joint probability density function of X and Y is given by
$\displaystyle
f(x,y) = \begin{cases} exp[-(x+y)] \qquad 0 \le x < \infty \; , \; 0
\le y < \infty \\
0 \qquad , otherwise
\end{cases}
$
Compute E[X] and P[Y>2]
As with most of the questions you have asked, apply the basic definitions:
$\displaystyle E(X) = \int_{0}^{+ \infty} \int_{0}^{+ \infty} x e^{-(x+y)} \, dy \, dx$.
$\displaystyle \Pr(Y > 2) = \int_{2}^{+ \infty} \int_{0}^{+ \infty} e^{-(x+y)} \, dy \, dx$.
Now integrate (it is assumed that you have the necessary calculus background).
By inspection the two rvs are independent.
X is an exponential with mean 1, hence EX=1, same for Y.
Since $\displaystyle \int_0^{\infty} e^{-x} \, dx=1$ you only need to compute $\displaystyle \int_2^{\infty}e^{-y} \, dy =e^{-2}$.