# B The CDF of the Sum of Independent Random Variables

1. Aug 4, 2016

### EngWiPy

Hello all,

Suppose I have the following summation $X=\sum_{k=1}^KX_k$ where the $\{X_k\}$ are independent and identically distributed random variables with CDF and PDF of $F_{X_k}(x)$ and $f_{X_k}(x)$, respectively. How can I find the CDF of $X$?

2. Aug 4, 2016

3. Aug 4, 2016

### EngWiPy

Can I use the Laplace transform instead of the Fourier transform. I now remember in the past using the moment generating function (MGF). So, can I find the MGF of $X$, and then find the inverse Laplace transform to find the CDF?

Last edited: Aug 4, 2016
4. Aug 5, 2016

### mathman

You can use the Laplace transform as you described. The expressions for the Laplace and Fourier transforms are similar. My personal preference is Fourier, since the inverse transform seems easier.

5. Aug 5, 2016

### chiro

He S_David.

I should point out that if the random variables are discrete random variables (as opposed to continuous ones) then you should look into probability generating functions.

6. Aug 5, 2016

### EngWiPy

Thanks. They are actually continuous random variables.

7. Aug 6, 2016

### mathman

One additional advantage to Fourier transform as opposed to Laplace is when moments are infinite. Example - Cauchy distribution.

8. Aug 6, 2016

### EngWiPy

The MGF of a random variable X is defined as

$$\mathcal{M}_X(s)=E\left[e^{-sX}\right]=\int_Xe^{-sX}f_X(x)\,dx$$

In this definition we have the PDF of X is involved in the definition of the MGF. How the Fourier transform is similar to this? I mean, in the definition of the Fourier transform there is not PDF involved, right? How to get the CDF from the Fourier transform then?

9. Aug 6, 2016

### chiro

10. Aug 7, 2016

### mathman

The limits of integration for a moment generating function are (in general) $(-\infty ,\infty )$, so that it is possible the integral may not exist.

The Fourier transform is that of the PDF (similar to Laplace, except using $e^{isx}$). To get CDF from Fourier transform, get PDF (using inverse transform) and integrate.

11. Oct 9, 2016

### bpet

There is a more general formula for when the variable can be either discrete, continuous, or a mixture of the two (or even singular if you wish).

We have
$$P[X+Y\le x] = P[X\le x-Y] = E[F_X(x-Y)] = \int F_X(x-y)dF(y)$$

The integral above is a Stieltjes integral so we recover the standard convolution formulas when X and Y are both purely discrete or purely absolutely continuous.