Question

The joint probability density function of X and Y is given by

f(x,y) = $\displaystyle \begin{cases} exp[-(x+y)] & 0 \le x < \infty , 0 \le y < \infty \\

0 & otherwise \end{cases}$

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

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- Dec 17th 2009, 07:14 PMzorroJoint probability density function
Question

The joint probability density function of X and Y is given by

f(x,y) = $\displaystyle \begin{cases} exp[-(x+y)] & 0 \le x < \infty , 0 \le y < \infty \\

0 & otherwise \end{cases}$

Compute E[X] and P[Y>2] - Dec 17th 2009, 08:35 PMmatheagle
They are indep by inspection.

Both X and Y are exponential with mean 1.

So E(X)=1 and

$\displaystyle P(Y>2)=\int_2^{\infty}e^{-y}dy=e^{-2}$ - Dec 17th 2009, 09:51 PMzorroI am stuck

**Here is what i have done**

g(x) = $\displaystyle \int_{0}^{ \infty}$ f(x,y) dy

= $\displaystyle \int_{0}^{ \infty} $ exp[-(x+y)] = -$\displaystyle \int_{0}^{ \infty}$ exp(x+y) dy

= - $\displaystyle \int_{0}^{ \infty}$ $\displaystyle exp(x) \ . \ exp(y) dy$

= -exp(x) $\displaystyle \int_{0}^{ \infty}$ exp(y) dy = ?_________I am stuck here what is exp(y).... - Dec 17th 2009, 10:04 PMmatheagle
INFINITY since you forgot the negative in the exponent.

DID you read my previous post? - Dec 17th 2009, 10:23 PMzorroI m stuck
- Dec 17th 2009, 10:28 PMmatheagle
I'll fix this, but there is to reason to do it this way...

$\displaystyle g(x) = e^{-x}\int_{0}^{ \infty}e^{-y}dy=e^{-x} $

you should know that the derivative and integral of $\displaystyle e^{-y}$ is $\displaystyle -e^{-y}$ - Dec 18th 2009, 12:48 AMzorroNow i am again stuck
- Dec 18th 2009, 02:03 AMmr fantastic
Why are you confused? Matheagle's post says clearly that $\displaystyle g(x) = e^{-x}$. You should be able to calculate $\displaystyle \int_{0}^{ \infty}e^{-y} \, dy = 1$ .... (Besides, a pdf cannot be negative and $\displaystyle -e^{x} < 0$ for all real values of x ....!!)

- Dec 18th 2009, 02:16 AMzorroThank u so much