# Convolution of continuous and discrete distributions

Assume we have two random variables $X$ and $Y$, such that $X \sim P(x)$ and $Y \sim G(y)$.

We ask, what is the distribution of $Z = X+Y$.

If both of the distributions of $X$ and $Y$ are discrete, the distribution of $Z$ is given by the convolution of the two, ie. $$F(z) = \sum_{x \in \mathbb Z} P(x) G(z-x).$$

Similarly, if the distributions are continuous, the convolution is $$F(z) = \int_{-\infty}^{\infty} P(x) G(z-x) dx.$$

Now, what if other, say $P(x)$ is discrete and the other continuous?

Take as an example probably the simplest one, $X \sim \text{Bernoulli}(1/2;x)$ and $Y \sim U(0,1)$.

If one thinks of this, $Z$ has probability $1/2$ of being between zero and 1 and same probability of being between one and two. Thus one could say immediately that $Z \sim U(0,2).$

The way I have seen this done elsewhere involves the use of a distribution called Dirac delta function $\delta(x)$, which is zero everywhere else except at zero but the integral over all reals is $1$.

Using this, we can make $\text{Bernoulli}(1/2;x)$ a continuous distribution, namely $$\text{DeltaBernoulli(1/2;x)} \sim \text{Bernoulli}(1/2;x)\delta(x)+\text{Bernoulli}(1/2;x)\delta(x-1).$$ (Subquestion: is this convolution of the two distributions or something else?)

Using this, lets perform the convolution $$M(z) = \int_{-\infty}^{\infty} 1*\text{PDF}(\text{DeltaBernoulli}(1/2;z-x)) dx = \begin{cases} 1/2 & 0\le z< 1 \vee 1 < z \le 2 \\ 1 &z = 1 \\ 0 &\text{elsewhere} \end{cases}$$

This is almost $U(0,2)$ and in any case has no real differences to it.

This method seems a tad clumsy, especially if one would want to convolute, say, normal distribution and Poisson distribution.

Are there other ways to do this?

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{{\rm F}\pars{z} = \int_{-\infty}^{\infty}{\rm P}\pars{x}{\rm G}\pars{z-x} \,\dd x}$

If ${\rm P}\pars{x}$ is discrete, we can write it as $\ds{{\rm P}\pars{x} = \sum_{n}P_{n}\delta\pars{x - x_{n}}}$ where $\ds{\braces{P_{n}}}$ is the probability of $x_{n}$. Then, $$\color{#0000ff}{\large% {\rm F}\pars{z} = \int_{-\infty}^{\infty}{\rm P}\pars{x}{\rm G}\pars{z-x}\,\dd x = \sum_{n}P_{n}\,{\rm G}\pars{z - x_{n}}}$$

If the random variables are independent, you can use the characteristic functions of the Random variables since if:

$$Sn=\sum_{i=1}^{n}a_iX_i$$

then

$$\phi_{S_n}(t)=\prod_{i=1}^n\phi_{X_i}(a_it)$$

So for the case given

$$\phi_{X}=1-\frac{1}{2}+\frac{1}{2}e^{it} \text{ and } \phi_Y=\frac{e^{it1}-e^{it0}}{it(1-0)}$$

so $$\phi_Z=\frac{e^{2it}-e^{it0}}{it(2-0)}$$ which is $U(0,2)$, not just close to.