Distribution of Sum of Two Weird Random Variables....

[Steve Zissou]

Hi there.

Let's say I have the following relationship:

x = a + b*z + c*y

z is distributed normally
y is distributed according to a different distribution, say exponential

Is there a way to figure out what is the distribution of x?

Thanks!

 

[FactChecker]

I will assume that the two random variables are independent.
The constant offset, $a$, can always be dealt with last. So let's ignore it for now.
Consider the sum of two random variables, $r_1 = b*z$ and $r_2 = c*y$ with distributions $p_1(r_1)$ and $p_2(r_2)$, respectively.
The distribution, $p(x)$, of the sum, $x = r_1+r_2$ is the convolution, $p(x) = \int_{t=-\infty}^{t=\infty}p_1(t)p_2(x-t) \,dt$

 

[Steve Zissou]

Nice! Thank you very much, FactChecker!

 

[mjc123]

For a given value x = X,
What is the probability that z has a value Z?
What is the probability that y has the value Y = (X -a -bZ)/c?
Integrate over z: f(x=X) = ∫f(z=Z)f(Y=(X-a-bZ)/c)dZ
(Assuming z and y are distributed independently. If not, you have to use a conditional probability for Y.)
Note that the range of z and y may be limited to less than their full possible range, e.g. if z is normal, it can take negative values, but if y is exponential it can only be positive (or zero). Therefore Z is limited to values for which X - a - bZ is nonnegative, ie Z ≤ (X - a)/b. (That's if b and c are both positive, work it out for yourself for other cases.)

 

[Steve Zissou]

Thanks, mjc123!