# Thread: A continous random variable...

1. ## 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]

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

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

3. ## I am stuck

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)....

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

5. ## I m stuck

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 ............

6. 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 $\displaystyle -e^{-y}$

7. ## Now i am again stuck

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?????

8. 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 ....!!)

9. ## Thank u so much

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