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Let $X$ be a discrete random variable with finite expectation, and let $a,b∈R$ whit $a≠0$.

Prove that the discrete random variable $Y=aX+b$ :

  • holds for expectation $\mathsf E(Y)=a⋅\mathsf E(X)+b$,
  • and for variance $\mathsf {Var}(Y)=a^2⋅\mathsf {Var}(X)$.

Suggestion: Consider the following theorem:

if $X$ is a discrete random variable, with $R_X$ rank and probability function $f_X$, and $g:R_X→R$ is a (measurable) function, then $\displaystyle \mathsf E(g(X))=\sum_{x\in R_X}~g(x)\cdot f_X(x)$.

The main problem is that I know what the theorem says, but I do not know how to use it. That is, I really do not know what values take $ g (x) $ nor $ f_X (x) $.

Could you explain the theorem and how would you use it for this demonstration?

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    To compute $E(Y)$, apply the theorem with $$g:x\mapsto ax+b$$ To compute $E(Y^2)$ (and deduce the variance of $Y$ from $E(Y^2)$ and $E(Y)$), apply the theorem with $$g:x\mapsto (ax+b)^2$$ – 2017-02-11

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In this situation $Y : \Omega \to \mathbb{R}$ is a composition $g(X)$ where $g$ is a measurable function $g : \mathbb{R} \to \mathbb{R}$. Plugging $Y = g(X)$ for $g(x) = ax + b$ in the definition you gave, \begin{align*} E(Y) &= \sum_{x\in R_X}(ax + b)f_X(x) \\ &= a \sum_{x \in R_X} xf_X(x) + b\sum_{x \in R_X}f_X(x) \\ &= aE(X) + b \end{align*} For the variance, we have by definition \begin{align*} Var(Y) &= E[(Y-EY)^2] \\ &= E[(aX+b - aEX - b)^2] \\ &= E[(aX - aEX)^2] \\ &= a^2 E[(X - EX)^2] \\ &= a^2 Var(X). \end{align*}

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    Maybe the second part is... \begin{align*} Var(Y) &= E[(Y-E(Y))^2] \\ &= E[(aX+b - aE(X) - b)^2] \\ &= E[(aX - aE(X))^2] \\ &= a^2 E[(X - E(X))^2] \\ &= a^2 Var(X). \end{align*} like this? – 2017-02-10
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    "In this situation $Y$ is your measurable function $g : R_{X} \to R$" Sorry but this is so wrong... No, $Y$ is not $g$, actually $Y=g(X)$ for some well chosen $g:\mathbb R\to\mathbb R$. – 2017-02-11
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    Carlos Frostte, is your question whether or not $E(X) = EX$? Yes that is what I meant. @Did Sorry I wasn't rigorous. I was trying to give a quick (but stupid) intuitive way to view the problem. – 2017-02-12