# Show that the sum of two random variables is a mixture

Take any $$({\lambda},{\mu},F,G)$$ such that

1) $$\lambda\equiv (\lambda_1,..., \lambda_J)$$, $$\lambda_j\in (0,1)$$ for each $$j=1,...,J$$ and $$\sum_{j=1}^J \lambda_j=1$$

2) $$\mu\equiv (\mu_1,..., \mu_J)$$ with $$\mu_1<...<\mu_J$$.

3) $$F$$ is a cumulative distribution function with steps of height $$\lambda_j$$ at each $$\mu_j$$.

4) $$G$$ is a cumulative distribution function whose associated probability mass function of probability density function is symmetric around zero.

Consider the mutually independent random variables, $$Y\stackrel{d}{\sim} F$$ and $$Z\stackrel{d}{\sim} G$$.

Question: Show that the cumulative distribution function of $$Y+Z$$ is the mixture $$\sum_{j=1}^{J} \lambda_j G(x-\mu_j)$$ at each $$x \in \mathbb{R}$$.

I have done some simulations and realised that the result seems in fact to hold. However, when I try to prove it formally, I'm completely stuck. Could you help? Even some informal intuition would be very useful.

• Simply convolve the distributions. Because convolution is a linear operator, a mixture input becomes a mixture output. This is practically a tautology, but its basic simplicity suggests that a mindless application of the definitions will take you directly to your goal. – whuber Jun 5 '20 at 14:07
• This is my attempt following your suggestion. Consider for simplicity the case where $Y$ and $Z$ are discrete. If $Y$ and $Z$ are independent, then the probability mass function of $X\equiv Y+Z$ evaluated at $x$ is $\sum_{l=-\infty}^\infty Pr(Y=l)Pr(Z=x-l)=\sum_{j=1}^J \lambda_j g(x-\mu_j)$, where $g$ is the probability mass function associated with the CDF $G$. In turn, the CDF of $X$ evaluated at $x$ is $\sum_{j=1}^J \lambda_j G(x-\mu_j)$. – TEX Jun 5 '20 at 14:11
• You can do that. However, your approach ignores the essential simplicity of the situation. I think you can obtain better insight by considering the more general situation where $F$ is a mixture of arbitrary distributions and $G$ is also an arbitrary distribution. Then (adopting a suggestive meta-notation), you can write a mixture as $F=\sum_i \pi_iF_i$ and, using $\star$ for the distribution arising from summing the underlying random variables, compute $$F\star G = \left(\sum_i \pi_iF_i\right)\star G = \sum_i\pi_i\left(F_i\star G\right),$$ QED. – whuber Jun 5 '20 at 14:20
• How can a CDF be symmetric about $0$? CDFs are non decreasing functions. – Dilip Sarwate Jun 5 '20 at 19:39
• @Dilip One need only adopt a relevant definition of "symmetric." For instance, the CDF of a random variable $X$ symmetric about a value $\mu$ satisfies $F_X(\mu+x)=1-F_X(\mu-x)$ for all real numbers $x$ where either of $\mu\pm x$ is a point of continuity. – whuber Jun 6 '20 at 17:52

## 1 Answer

This ultimately is Fubini's Theorem, but to keep the analysis elementary let's stick to a finite mixture of arbitrary distributions. By definition, this means $$Y$$ can be considered in terms of other variables $$Y_j$$ with cumulative distribution functions (CDFs) $$F_j$$ and that for any number $$y,$$

$$F(y) = \Pr(Y \le y) = \sum_{j} \lambda_j \Pr(Y_j \le y) = \sum_j \lambda_j F_j(y).$$

Now let $$x$$ be any number and compute the CDF of $$Y+Z$$ at $$x$$ in terms of the CDF $$G$$ of $$Z$$ (defined by $$G(z) = \Pr(Z \le z)$$) as

\eqalign{ {\Pr}_{Y,Z}(Y+Z \le x) &= {\Pr}_{Y,Z}( Z \lt x - Y ) \\ &= \mathbb{E}_Y (G(x - Y)) \\ &= \int G(x-y)\,\mathrm{d}F(y) \\ &= \int G(x-y)\, \mathrm{d}\left(\sum_j \lambda _j F_j(y)\right) \\ &= \sum_j \lambda_j \int G(x-y)\,\mathrm{d}F_j(y).\tag{*} }

(The switching of the order of integration and summation merely expressed the linearity of integration but can be seen as an example of Fubini's Theorem.)

In the question itself, the component distributions are atoms at the $$\mu_j$$ and their CDFs $$F_j,$$ which jump from a value of $$0$$ to a value of $$1$$ at $$\mu_j,$$ have the property that for any piecewise differentiable function $$H$$ with $${\lim}_{x\to\infty}H(x)=0,$$

\eqalign{ \int H(y)\,\mathrm{d}F_j(y) &= H(y)F_j(y)\mid_{-\infty}^\infty - \int H^{\prime}(y) F_j(y)\,\mathrm{d}y \\ &=(0 - 0) - \left(\int_{-\infty}^{\mu_j}H^\prime(y)(0)\mathrm{d}y + \int_{\mu_j}^\infty H^\prime(y)(1)\mathrm{d}y\right) \\ &= (0 - 0) - \left(0 + (0 - H(\mu_j))\right) \\ &= H(\mu_j). }

The integrals are all in the sense of Riemann or Lebesgue and the initial equality is integration by parts. The zeros arise from the assumed limiting value of $$H$$ at $$\infty$$ and from the fact that $$F_j$$ is identically zero for very negative arguments.

Consequently, applying this to the function $$H: y \to G(x-y)$$ (whose limit as $$y\to\infty$$ clearly is zero), the general result $$(*)$$ reduces to

$$\Pr(Y + Z \le x) = \sum_j \lambda_j H(\mu_j) = \sum_j \lambda_j G(x-\mu_j),$$

QED.

Note $$G$$ does not have to be symmetric, but it is important that it be piecewise differentiable. This includes the CDFs of continuous random variables, the usual discrete random variables (with no accumulation points among their supports), and mixtures thereof.