# Prove that $\lim_{c \to \infty} \text E(X\chi_c) = \lim_{c \to \infty}\text E(\lvert X \rvert \chi_c) = 0$

Edit:

I was told to use the Dominated Convergence Theorem instead, so below I try that.

My 2nd Attempt (DCT):

Let $$(c_n)_{n = 1}^{\infty}$$ be a monotonic sequence going to infinity. This gives us a sequence of random variables $$\{ X\chi_{c_n} \}_{n = 1}^{\infty}$$. For a dominating function, we will use $$\lvert X \rvert$$. This works because we are told that $$\text E(\lvert X \rvert) < \infty$$, and

\begin{align*} \lvert X\chi_{c_n} \rvert &= \lvert X \rvert \lvert \chi_{c_n} \vert\\ &= \lvert X \rvert \chi_{c_n}\\ &\leq \lvert X \rvert. \end{align*}

The sequence of random variables $$\{ X\chi_{c_n} \}_{n = 1}^{\infty}$$ converges to the zero random variable. (Is that right?) According to the DCT, therefore,

\begin{align*} \lim_{n \to \infty} \text E(X\chi_{c_n}) &= \text E(0)\\ &= 0\\ \implies \lim_{c \to \infty} \text E(X\chi_c) &= 0. \end{align*}

Assuming this is correct, I still need to think about how to show that $$\lim_{c \to \infty} \text E(\lvert X \rvert\chi_c) = 0$$. Maybe I could just apply the DCT a second time, to $$\{ \lvert X \rvert \chi_{c_n} \}_{n = 1}^{\infty}$$.

Question:

Let $$X$$ be a real-valued random variable on a probability space $$(\Omega, \mathscr F, P)$$ for which $$\text E(\lvert X \rvert) < \infty$$. For $$c \geq 0$$, let $$\chi_c$$ be the indicator function for the event $$\{ \omega \colon X(\omega) \geq c \}$$. Prove that $$\lim_{c \to \infty} \text E(X\chi_c) = \lim_{c \to \infty}\text E(\lvert X \rvert \chi_c) = 0$$.

My attempt:

I have no idea what I'm doing, but I thought the Cauchy-Schwarz inequality might be applicable, so I started noodling around with that.

For random variables $$Y$$ and $$Z$$ meeting certain conditions, the Cauchy-Schwarz inequality says that $$\text E(YZ)^2 \leq \text E(Y^2)\text E(Z^2).$$

So I consider the random variables $$\lvert X \rvert$$ and $$\chi_c$$. We have

\begin{align*} \text E(\lvert X \rvert \chi_c)^2 &\leq \text E(\lvert X \rvert^2) \text E(\chi_c^2)\\ &= \text E(X^2)\text E(\chi_c)\\ &= \text E(X^2)P(X \geq c). \end{align*}

Take the limit of both sides as $$c \to \infty$$, and $$P(X \geq c)$$ should go to zero, so then $$\text E(\lvert X \rvert \chi_c)^2$$ goes to zero. Therefore $$\text E(\lvert X \rvert \chi_c)$$ goes to zero.

Since $$X \chi_c \leq \lvert X \rvert \chi_c$$, we must have $$\text E(X \chi_c) \leq \text E(\lvert X \rvert \chi_c)$$, so $$\text E(X\chi_c)$$ goes to zero.

I'm not sure if any of this is rigorous enough. Everything after "Take the limit..." above seems hand-wavey.

For example:

• I need finite second moments to apply the Cauchy-Schwarz inequality, but I don't see where this assumption is valid. (I guess the characteristic function, as a Bernoulli random variable, must have a finite second moment.)

• I feel like $$P(X \geq c)$$ should go to zero but I'm not sure how to prove it.

• If $$\text E(\lvert X \rvert \chi_c)^2$$ goes to zero, then does $$\text E(\lvert X \rvert \chi_c)$$ go to zero? I guess it would by some kind of continuity argument?

• If everything above in this list is correct, then I think the rest of the argument should work. Is that right?

Thanks for any help.

• "I need finite second moments to apply the Cauchy-Schwarz inequality". Precisely. And the second moment can be infinite. You need to use the dominated convergence theorem. Commented Mar 14, 2022 at 20:50
• @zhoraster Thanks. I need to go re-read the DCT and figure out how I would apply it here. Would the sequence of measurable functions be $X\chi_c$? And then the dominating function would be $\lvert X \rvert \chi_c$? I need to think about this. Commented Mar 14, 2022 at 20:56
• The dominant is $|X|$. Commented Mar 14, 2022 at 20:58
• Maybe that doesn't make sense. But I figure I need a sequence of $c$ values going to infinity. (I am not great at analysis.) Commented Mar 14, 2022 at 20:58

Yes, the sequence $$(X\chi_{c_n})$$ converges pointwise to the zero random variable, which means that for every $$\omega\in\Omega$$ the sequence of real numbers $$(X\chi_{c_n}(\omega))$$ converges to zero. Why is this true? Argue as follows: Let $$\omega\in\Omega$$. Then the value $$X(\omega)$$ is a real number. That real number will be less than $$c_n$$ when $$n$$ gets big enough, hence the indicator $$\chi_{c_n}(\omega)$$ will be zero for all large $$n$$, hence the random variable $$X\chi_{c_n}$$ has value $$X(\omega)\chi_{c_n}(\omega)=0$$ for all large $$n$$.
As for proving that $$E(|X|\chi_{c})\to0$$, there's no need to apply DCT again. Just observe that $$|X|\chi_{c}$$ is the same random variable as $$X\chi_{c}$$ when $$c>0$$. Can you see why?
• Thanks. I'm still thinking about this, but I need to consider $c \geq 0$, not $c > 0$. I'm trying to figure out if that's important or not. Commented Mar 14, 2022 at 21:58
• The distinction is not important. If $c_n\to\infty$, then $c_n$ will be strictly positive after a certain point. Commented Mar 14, 2022 at 22:03
• I think that $\lvert X \rvert \chi_c$ and $X \chi_c$ are the same for $c \geq 0$ because for an $\omega$ such that $X(\omega) < c$, they're both zero, and for an $\omega$ such that $X(\omega) \geq c$, they're both equal to $X(\omega)$. Thanks for your help. Commented Mar 15, 2022 at 0:26