Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.