Let $X_0, X_1, X_2, ..., X_n$ each be non-identical independent random variables.

Let $x_0, x_1, ... , x_n$ be possible values of each of those random variables.

Let $\newcommand{\Pdf}{\operatorname {Pdf}} \Pdf_{X_0}(x_{0}), \Pdf_{X_1}(x_{1}), ... , \Pdf_{X_n}(x_{n})$ represent each of random variables corresponding probability density functions.

$X_0$ has probability density function $\Pdf_{X_0}(x_0)$ ,

$X_1$ has probability density function $\Pdf_{X_1}(x_1)$ ... etc...

Lets say we know explicitly, analytically, exactly what all the $Pdf$'s are above.

Now Let $Y = X_0 + X_1 + X_2, ... + X_n$ which has probability density function $\Pdf_Y(y)$.

What is $\Pdf_Y(y)$ ?

How do I calculate it with the information above?


It's a well known (and important) result that the density of the sum of independent random variables is the convolution of the individual densities.

See for example here or here or here.

Let $\tau_{0}, \tau_{1}, ... \tau_{n-1}$ represent temporary variables which are integrated out as part of the convolution.

Let $PDF_{x_{a} + x_{b}} $ denote the PDF of only two random variable $X_a + X_b$

$PDF_{x_{a} + x_{b}}(t) = PDF_{x_{a}} * PDF_{x_{a}} (t) = \int_{-\infty}^{\infty} PDF_{x_{a}}( \tau ) PDF_{x_{a}}( t - \tau ) d\tau $

Then the following follows:

$PDF_y(y) = PDF_{x0}*PDF_{x1}*PDF_{x2}*...*PDF_{xn}(y)$ $= \int_{-\infty}^{\infty}...\int_{-\infty}^{\infty} PDF_{x0}(\tau_{0}) PDF_{x1}(\tau_{1} - \tau_{0}) PDF_{x2}(\tau_{2} - \tau_{1} ) ... PDF_{xn}(y - \tau_{(n-1)}) d\tau_{0} d\tau_{1} ... d\tau_{(n-1)} $


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.