# Compute E[X] and P[Y>2]

• May 4th 2009, 05:32 AM
Stats
Compute E[X] and P[Y>2]
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]
• May 4th 2009, 03:44 PM
mr fantastic
Quote:

Originally Posted by Stats
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).
• May 4th 2009, 05:00 PM
matheagle
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 $\displaystyle \int_2^{\infty}e^{-y} \, dy =e^{-2}$.