A continous random variable...

**zorro** · December 17th 2009, 07:14 PM

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

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

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

**matheagle** · December 17th 2009, 08:35 PM

They are indep by inspection.
Both X and Y are exponential with mean 1.
So E(X)=1 and

$P(Y>2)=\int_2^{\infty}e^{-y}dy=e^{-2}$

**zorro** · December 17th 2009, 09:51 PM

Originally Posted by matheagle

They are indep by inspection.
Both X and Y are exponential with mean 1.
So E(X)=1 and

$P(Y>2)=\int_2^{\infty}e^{-y}dy=e^{-2}$

Here is what i have done

g(x) = $\int_{0}^{ \infty}$ f(x,y) dy
= $\int_{0}^{ \infty}$ exp[-(x+y)] = - $\int_{0}^{ \infty}$ exp(x+y) dy
= - $\int_{0}^{ \infty}$ $exp(x) \ . \ exp(y) dy$

= -exp(x) $\int_{0}^{ \infty}$ exp(y) dy = ?_________I am stuck here what is exp(y)....

**matheagle** · December 17th 2009, 10:04 PM

INFINITY since you forgot the negative in the exponent.
DID you read my previous post?

**zorro** · December 17th 2009, 10:23 PM

Originally Posted by matheagle

INFINITY since you forgot the negative in the exponent.
DID you read my previous post?

I have read ur prev post, but what about the -ve sign ..........i have not forgotten about it but i feel i am missing something ............

**matheagle** · December 17th 2009, 10:28 PM

I'll fix this, but there is to reason to do it this way...

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

you should know that the derivative and integral of $e^{-y}$ is $-e^{-y}$

**zorro** · December 18th 2009, 12:48 AM

Originally Posted by matheagle

I'll fix this, but there is to reason to do it this way...

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

you should know that the derivative and integral of $e^{-y}$ is $-e^{-y}$

g(x) = $e^{-x}$ or $-e^x$ .......which one is it?????

**mr fantastic** · December 18th 2009, 02:03 AM

Originally Posted by zorro

g(x) = $e^{-x}$ or $-e^x$ .......which one is it?????

Why are you confused? Matheagle's post says clearly that $g(x) = e^{-x}$ . You should be able to calculate $\int_{0}^{ \infty}e^{-y} \, dy = 1$ .... (Besides, a pdf cannot be negative and $-e^{x} < 0$ for all real values of x ....!!)

**zorro** · December 18th 2009, 02:16 AM

Originally Posted by mr fantastic

Why are you confused? Matheagle's post says clearly that $g(x) = e^{-x}$ . You should be able to calculate $\int_{0}^{ \infty}e^{-y} \, dy = 1$ .... (Besides, a pdf cannot be negative and $-e^{x} < 0$ for all real values of x ....!!)

Thanks u so much Mr fantastic and matheagle.........Zorro will never forget it .........Its because of u i am able to wave my sword in the air ....On gourd.......
cheers mate

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