4
$\begingroup$

Let $(\Omega,\mathcal{F},P)$ denote the probability space on which all of the following random variables are defined and $\omega\in\Omega$. Let $X$ denote a non-negative random variable and $X_n,n\geq 1$ a random variable which is defined as follows: $$X_n (\omega) = \sum_{k=1}^{n2^n} \frac{k - 1}{2^n}\cdot\mathbf{1}_{X (\omega)\in\left[\frac{k - 1}{2^n}, \frac{k}{2^n}\right)} + n\cdot\mathbf{1}_{X (\omega)\in \left[n,\infty\right)}.\tag{$\ast$}$$

I see that $X_n$ is simple. I want to find its expected value. In Billingsley (1995): Probability and Measure it says on p. 68 (Equation (5.2)) that a simple random variable $Y$ has the form $$Y (\omega) = \sum_i y_i \mathbf{1}_{\omega\in A_i},\tag{$\ast\ast$}$$ where the $y_i$ are the values taken by $Y$ and the $A_i$ form a partition of $\Omega$. On p. 76 (Equation (5.15)) the expected value of a simple random variable in the form $(\ast\ast)$ is given by $$E\left[Y\right] = \sum_i y_i P(A_i).$$

My understanding is that $(\ast)$ is not in the form $(\ast\ast)$ because of $X (\omega)$ instead of $\omega$ as argument of the indicator functions. Is that correct? How can I obtain $(\ast)$ in the form $(\ast\ast)$? Or is that not a path to the expected value of $(\ast)$?

$\endgroup$
2
  • 1
    $\begingroup$ $(*)$ is already of the form of $(**)$. Notice that $\boldsymbol 1_{X(\omega )\in [a,b)}$ means $\boldsymbol 1_{\{\omega \mid X(\omega )\in [a,b)\}}=\boldsymbol 1_{X^{-1}([a,b)}$. $\endgroup$
    – Surb
    Commented Mar 23, 2020 at 13:10
  • $\begingroup$ I see your point that $\mathbf{1}_{X (\omega)\in\left[a,b\right)}$ is the same as $\mathbf{1}_{\omega\in\left\{\omega | X (\omega)\in\left[a,b\right)\right\}}$. I don't get the RHS of the equation though. It looks to me as if you pass an interval to $X^{-1}$. But $X (\omega)$ gives a real-valued scalar, so how can you pass a set to it? $\endgroup$
    – nluckn
    Commented Mar 23, 2020 at 14:21

1 Answer 1

3
$\begingroup$

Use the linearity of expectations, i.e., \begin{align} \mathsf{E}X_n &= \mathsf{E}\!\left[\sum_{k=1}^{n2^n} \frac{k - 1}{2^n}\cdot\mathbf{1}_{X \in\left[\frac{k - 1}{2^n}, \frac{k}{2^n}\right)} + n\cdot\mathbf{1}_{X\in \left[n,\infty\right)}\right] \\ &=\sum_{k=1}^{n2^n} \frac{k - 1}{2^n}\cdot\mathsf{P}\!\left(X \in\left[\frac{k - 1}{2^n}, \frac{k}{2^n}\right)\right) + n\cdot\mathsf{P}(X\in \left[n,\infty\right)). \end{align}

$\endgroup$
2
  • $\begingroup$ So are $Y (\omega) = \sum_i y_i\mathbf{1}_{Z (\omega)\in A_i}$ and $E\left[Y\right] = \sum_i y_i P (Z (\omega)\in A_i)$ with $Z$ another random variable generalizations of $(\ast\ast)$ and the third equation and it's those which you use? $\endgroup$
    – nluckn
    Commented Mar 23, 2020 at 13:53
  • $\begingroup$ The correct formula for $Y$ taking values in $\{y_1,\ldots,y_m\}$ is $$ \mathsf{E}Y=\sum_{i=1}^m y_i \mathsf{P}(Y=y_i). $$ In your case, $Y=X_n$, the event $\{Y=y_i\}=\{X\in \ldots\}$. $\endgroup$
    – user140541
    Commented Mar 23, 2020 at 15:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .