# Thread: Function of random variables

1. ## Function of random variables

can some one help in this question:

If X is Normal($\displaystyle \mu,$$\displaystyle \sigma^2) and Y is Exponentioal (\displaystyle \lambda) What is the distribution for: Z=X+Y Z=X-Y Z=X*y Z=x/Y ineed these distribution quickness please 2. Originally Posted by doaa can some one help in this question: If X is Normal(\displaystyle \mu,$$\displaystyle \sigma^2$) and Y is Exponentioal ($\displaystyle \lambda$)
What is the distribution for:

Z=X+Y
Z=X-Y
Z=X*y
Z=x/Y

ineed these distribution quickness

Are X and Y independent?

3. Yes they independent

4. Originally Posted by doaa
Yes they independent
Then can't you write down the joint pdf of X and Y and then calculate the cdf of Z in each case.

(You could also use moment generating functions for Z = X + Y but then you'd need to identify the resulting mgf ....)

5. This is a nasty problem. There is no nice answer, i.e., nice form.
IF BOTH X and Y are normal or BOTH are exponential then you can use the MGF. BUT the MGF here won't be recognizable otherwise and that only works for the sum and difference too. So that's worthless. I would just do a 2-2 tranform and integrate out the dummy variable. BUT I wonder if the instructor wanted both of the same form.

6. First thank you

But my doctor give me as homework and i tried to solve it but the formulas is complicated
and i though that my work is wrong
and i dont know what to do

7. Instead of 2-2 transform, it's probably easier to find $\displaystyle F_Z(z)$.

i) X + Y
$\displaystyle F_Z(z) = \int_{-\infty}^{\infty} \int_{-\infty}^{z-y}f_X(x)f_Y(y) dx dy$

ii) X-Y
$\displaystyle F_Z(z) = \int_{-\infty}^{\infty} \int_{-\infty}^{z+y}f_X(x)f_Y(y) dx dy$

iii) X*Y
$\displaystyle F_Z(z) = \int_{-\infty}^{\infty} \int_{z/y}^{0}f_X(x)f_Y(y) dx dy + \int_{-\infty}^{\infty} \int_{0}^{z/y}f_X(x)f_Y(y) dx dy$

iv) X/Y
$\displaystyle F_Z(z) = \int_{-\infty}^{\infty} \int_{-\infty}^{yz}f_X(x)f_Y(y) dx dy$

The upper and lower limit might not be correct but I hope you get the idea.