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

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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

$<br /> <br /> f(x,y) = \begin{cases} exp[-(x+y)] \qquad 0 \le x < \infty \; , \; 0<br /> \le y < \infty \\<br /> 0 \qquad , otherwise<br /> \end{cases}<br />$

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

$<br /> <br /> f(x,y) = \begin{cases} exp[-(x+y)] \qquad 0 \le x < \infty \; , \; 0<br /> \le y < \infty \\<br /> 0 \qquad , otherwise<br /> \end{cases}<br />$

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

As with most of the questions you have asked, apply the basic definitions:

$E(X) = \int_{0}^{+ \infty} \int_{0}^{+ \infty} x e^{-(x+y)} \, dy \, dx$ .

$\Pr(Y > 2) = \int_{2}^{+ \infty} \int_{0}^{+ \infty} e^{-(x+y)} \, dy \, dx$ .

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 $\int_0^{\infty} e^{-x} \, dx=1$ you only need to compute $\int_2^{\infty}e^{-y} \, dy =e^{-2}$ .

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