# How do you sum PDF's of random variables?

I have a question asking me to determine the PDF of $L=X+Y+W$, where $X$, $Y$ and $W$ are all independent. $X$ is a Bernoulli random variable with parameter $p$, $Y \sim \mathrm{Binomial}(10, 0.6)$ and $W$ is a Gaussian random variable with zero mean and unit variance (meaning is is a standard normal random variable).

I know the PDF's of $X$, $Y$ and $W$ (sort of hard to type out, but I know them). Could I get some sort of hint as to how these are added together?

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It is easier to work with their characteristic functions. $\psi_L = \psi_X \psi_Y \psi_W$ – user17762 Oct 22 '12 at 20:51

The PDF of the sum of two independent variables is the convolution of the PDFs: $$f_{U+V}(x) = \left( f_{U} * f_{V} \right) (x)$$ You can do this twice to get the PDF of three variables.

By the way, the Convolution theorem might be useful.

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Thank you very much, I wasn't even sure how to get started on this. – Hoser Oct 22 '12 at 20:59
I'm not really sure how to convolve a Bernoulli and Binomial random variable... any way to get a pointer with that? – Hoser Oct 22 '12 at 21:15
You can directly apply the integral definition of the convolution, use the convolution theorem, or as Marvis pointed out use the characteristic functions $\$[1]. Whichever hethod you use, at the last step where you take the (inverse) fourier transform you may not be able to simplify it so your answer is an integral. [1]:en.wikipedia.org/wiki/… . – Navin Oct 22 '12 at 23:52

Let me get you started by doing $X + Y$.

Let $X \sim Ber(p)$ and $Y \sim Bin(10, 0.6)$. Let $\mu$ be the law of $X$ and $\nu$ be the law of $Y$. The law of $X + Y$ is given by the convolution

$$(\nu * \mu)(H) = \int_{\mathbb{R}} \nu(H-x)\mu(dx), \qquad H \subseteq \mathbb{R}.$$

$X$ and $Y$ are discrete and $X + Y$ can take values $0, 1, \ldots, 11$. Thus we specify $(\mu * \nu)(k)$ for $k = 0, 1, \ldots 11$.

$$(\nu * \mu)(k) = \int_{\mathbb{R}} \nu(k-x)\mu(dx) = \nu(k)(1-p) + \nu(k-1)p. \qquad k = 0, 1, \ldots, 11.$$

To get the first equality substitute $k$ for $H$ into the convolution formula. The second equality uses the fact that we know $\mu$ because $X$ is $Ber(p)$. We also know $\nu$ and so we can write

$$(\nu * \mu)(k) = {10 \choose k} 0.6^k 0.4^{n-k}(1-p) + {10 \choose k-1} 0.6^{k-1} 0.4^{n-(k-1)}p$$

You can check that the convolution has given a sensible formula for the distribution of $X + Y$. For example if $k = 10$ then

$$(\nu * \mu)(10) = 0.6^{10} (1-p) + 10 \cdot 0.6^{9} p.$$

$X + Y = 10$ if $X = 0$ and $Y = 10$ or if $X = 1$ and $Y = 9$ and because $X$ and $Y$ are independent these formulas can be verified.

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Thank you so much – Hoser Oct 23 '12 at 13:01