How do you sum PDF's of random variables?

Question

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?

Answer 1

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.

Answer 2

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

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

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

\begin{equation} (\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. \end{equation}

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

\begin{equation} (\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 \end{equation}

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

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

$X + Y = 10$ 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.