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

Question

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 $\{ \lvert X \rvert \chi_{c_n} \}_{n = 1}^{\infty}$.

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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.

Questions and comments:

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.

Answer 1

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?