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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

Approach:

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!

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  • $\begingroup$ 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

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

$$\mathbb{E}[X+Y]=\mathbb{E}[X]+\mathbb{E}[Y]$$

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

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

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  • $\begingroup$ 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
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    $\begingroup$ @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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