# Representation and expected value of a certain simple r.v.

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 $$(\ast)$$?

• $(*)$ is already of the form of $(**)$. Notice that $\boldsymbol 1_{X(\omega )\in [a,b)}$ means $\boldsymbol 1_{\{\omega \mid X(\omega )\in [a,b)\}}=\boldsymbol 1_{X^{-1}([a,b)}$.
– Surb
Commented Mar 23, 2020 at 13:10
• 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? Commented Mar 23, 2020 at 14:21

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}
• 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? Commented Mar 23, 2020 at 13:53
• 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 $\{Y=y_i\}=\{X\in \ldots\}$.