Random Variable/ Expectations

Question

Can you please help me on this question:

(a) In class we define the expectation of a random variable $X$ as $$\mathbb{E}(X):=\sum_{\omega\in\Omega}X(\omega)\mathrm{Pr}[\omega]\;,$$ where $\Omega$ is the sample space. In the notes, the expectation is defined to be $$\mathbb{E}(X):=\sum_{a\in\mathscr{A}}a\mathrm{Pr}[X=a]\;.$$ where $\mathscr{A}$ is the range of values that $X$ can take on. Show that the two definitions are equivalent.

(b) Given a random variable $X$ defined on a sample space $\Omega$, $Y=X^2$ is also a random variable defined on $\Omega$. Why? (Hint: Look review [sic] the definition of a random variable)

(c) Show, from the definition of the expectation, that $$\mathbb{E}(X^2)=\sum_{a\in\mathscr{A}}a^2\operatorname{Pr}[X=a].$$ (You can start with the definition in class or the definition in the notes, since you have already shown in part (a) that they are equivalent.)

(d) Generalize the result in part (c) to give an expression for $\mathbb{E}(f(X))$, where $f$ is an arbitrary function from $\mathfrak{R}$ to $\mathfrak{R}$. You can give your answer without proof.

Thanks!

Answer 1

Here's an almost complete solution to (a) with some justifications left out. $$\begin{align*} E(X) & = \sum_{a\in\mathscr{A}} a\cdot \text{Pr}[X=a]\\ & = \sum_{a\in\mathscr{A}} ~~a\left(\sum_{\omega\in[X=a]} \text{Pr}[\omega]\right)\\ & = \sum_{a\in\mathscr{A}} ~~~~\sum_{\omega\in[X=a]} a\cdot \text{Pr}[\omega]\\ & = \sum_{a\in\mathscr{A}}~~~~~ \sum_{\omega\in[X=a]} X(\omega)\text{Pr}[a]\\ & = \sum_{\omega\in\Omega} X(\omega)\text{Pr}[\omega] \end{align*} $$ where the last equality is true since the events $[X=a]~\forall a\in \mathscr{A}$ partions the sample space, $\Omega$. Thus, summing over all the outcomes is the same as summing over $\Omega$.

I leave the justification of the other equalities for you.

Answer 2

For (a): For each $a\in\mathscr{A}$ let $\Omega_a=\{\omega\in\Omega:X(\omega)=a\}$. How is $$\sum_{\omega\in\Omega_a}X(\omega)\mathrm{Pr}[\omega]$$ related to $a\mathrm{Pr}[X=a]$?

Added: For each $\omega \in \Omega_a$, $X(\omega)=a$, so $$\sum_{\omega\in\Omega_a}X(\omega)\mathrm{Pr}[\omega]=a\sum_{\omega\in\Omega}\mathrm{Pr}[\omega]\;;$$ if you can show that $$\sum_{\omega\in\Omega_a}\mathrm{Pr}[\omega] = \mathrm{Pr}[X=a]\;,$$ you’ll be done. Now, how do you define $\mathrm{Pr}[X=a]$?