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:




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.


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.