# Can I average c.d.f's for the average of draws from different distributions?

Let's say I have 3 random variables, $X$, $Y$, $Z$ with c.d.f.'s $F_X(x)$, $F_Y(y)$, and $F_Z(z)$. All are supported over $(0,1)$

I want to define $Z$ such that $Z = aX + (1-a)Y$ (i.e. a weighted average of $X$ and $Y$ where $a$ is just some constant).

Does is follow then that $F_Z(z) = aF_X(z) +(1-a)F_Y(z)$?

I am pretty sure this is not true, though it would be true that $\mathbb{E}Z = a\mathbb{E}X +(1-a)\mathbb{E}Y$ but I wanted to check. Thanks for your help!

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It is not true. The correct way to go about is \begin{align} F_Z(z) & = \mathbb{P}(Z \leq z) = \mathbb{P}(aX + (1-a)Y \leq z)\\ & = \int_x \mathbb{P}(aX + (1-a)Y \leq z \vert X=x) \mathbb{P}(X \in (x,x+dx))\\ & = \int_x \mathbb{P}(ax + (1-a)Y \leq z) \frac{dF_X(x)}{dx} dx\\ & = \int_x F_Y \left(\frac{z-ax}{1-a} \right) \frac{dF_X(x)}{dx} dx \,\,\,\,\,\, (\text{Assuming }a <1) \end{align}

As a counterexample to your claim, consider $x \sim U([-10,-9])$ and $y \sim U([9,10])$. Take $a = \frac12$. This means that $Z = aX+(1-a)Y = \dfrac{X+Y}2 \in \left[-\frac12,\frac12 \right]$.

Note that for $z \in \left[-\frac12,\frac12 \right]$, we have that $F_X(z) = 1$ and $F_Y(z) = 0$.

Hence, $aF_X(z) + (1-a) F_Y(z) = \dfrac12$, which is clearly wrong.

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