I'm working on proving that

$E(aY+bZ | X) = aE(Y|X) + bE(Z | X)$

Where X, Y and Z are discrete random variables. Where we assume that all their (joint/marginal) probability mass functions and expectations exist


So I figured I have to use linearity of expectations. What bugs me, is what to do with the conditional statement and the constants.

My thought is to use independence, so $p(y|x)= p(y)$

And the constants I thought I could move outside, since $E(a) = a$

Then I could proceed with the proof:

$E(X+Y) = \sum_x \sum_y (x + y)P_{XY}(x,y)$

$=\sum_x \sum_y x \cdot P_{XY}(x,y) + \sum_x \sum_y y \cdot P_{XY}(x,y) $

$= \sum_x x \cdot \sum_y P_{XY} (x,y) + \sum_y y \cdot \sum_x P_{XY} (x,y)$

$= \sum_x x \cdot P_{X}(x) + \sum_y y \cdot P_{Y}(y) $

$= E(X) + E(Y)$

My question is, I don't really know if I just can apply the use of independence like that and then proceed with the proof of $E(X + Y) = E(X) + E(Y)$?

Hope you can help!

  • $\begingroup$ You need to add your definition of $E[Y\mid X]$ for a helpful and correct answer. $\endgroup$ Commented Dec 16, 2021 at 11:11

1 Answer 1


independence is not stated in the question and not needed to prove what you are requested to do.


is valid also if the rv's are not independent. Your proof is quite correct (I amended your $E(X,Y)$ in $E(X+Y)$) but it is not exactly what you are asked to do.

To prove your statement, including conditional probability and constants, simply use the definition finding

$$\begin{align} \mathbb{E}[aY+bZ|X] & = \sum_y\sum_z(ay+bz)p(y,z|x)\\ & = a\sum_y\sum_zyp(y,z|x)+b\sum_y\sum_zzp(y,z|x)\\ & = a\sum_y yp(y|x)+b\sum_z zp(z|x)\\ &=a\mathbb{E}[Y|X]+b\mathbb{E}[Z|X] \end{align}$$

  • $\begingroup$ Thank you for your help! Is this proof/approach called something? So I can look it up and study it further. $\endgroup$
    – bestmate21
    Commented Dec 16, 2021 at 10:35
  • 1
    $\begingroup$ @bestmate21 : no, it is just a matter of applying expectation's definition $\endgroup$
    – tommik
    Commented Dec 16, 2021 at 10:37
  • $\begingroup$ This instead argues that $E[aY+bZ\mid X=x]=a E[Y\mid X=x]+bE[Z\mid X=x]$. The distinction needs to be made clear. $\endgroup$ Commented Dec 16, 2021 at 10:52
  • $\begingroup$ @StubbornAtom Are you reffering to tommik's answer or the problem in general? Could you elaborate? $\endgroup$
    – bestmate21
    Commented Dec 17, 2021 at 10:40

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