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

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• Steve Zissou

#### 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!

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!

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.)

Thanks, mjc123!