# A continous random variable...

• Dec 17th 2009, 07:14 PM
zorro
Joint 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 PM
matheagle
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 PM
zorro
I am stuck
Quote:

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

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

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 PM
matheagle
INFINITY since you forgot the negative in the exponent.
DID you read my previous post?
• Dec 17th 2009, 10:23 PM
zorro
I m stuck
Quote:

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 ............(Angry)
• Dec 17th 2009, 10:28 PM
matheagle
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 AM
zorro
Now i am again stuck
Quote:

Originally Posted by matheagle
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 $\displaystyle -e^{-y}$

g(x) = $\displaystyle e^{-x}$ or $\displaystyle -e^x$.......which one is it?????
• Dec 18th 2009, 02:03 AM
mr fantastic
Quote:

Originally Posted by zorro
g(x) = $\displaystyle e^{-x}$ or $\displaystyle -e^x$.......which one is it?????

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 AM
zorro
Thank u so much
Quote:

Originally Posted by mr 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 ....!!)

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 (Beer)