3
$\begingroup$

I am reading Larry Wasserman's All of Statistics and exercise 2 in chapter 6 asks for a proof that given sequence of random variables $ X_1, X_2, \dots $, show that $ X \xrightarrow{\text{QM}} b $ if and only if

$$ \begin{align} & \lim_{n \rightarrow \infty} \mathbb{E}(X_n) = b & \text{and } & & \lim_{n \rightarrow \infty} \mathbb{V}(X_n) = 0. \end{align} $$

I'm getting stuck proving the forward direction. I started by expanding the definition of quadratic mean convergence as follows. By assumption, we have $$ \lim_{n \rightarrow \infty} \mathbb{E}(X-b)^2 = 0. $$

And then by linearity of expectation we have, $$ \lim_{n \rightarrow \infty} \mathbb{E}(X-b)^2 = \lim_{n \rightarrow \infty} \mathbb{E}(X_n^2) - 2b\ \mathbb{E}(X_n) + b^2 = 0. $$

This is where I get stuck. It seems like we will somehow get that $ \mathbb{E}(X_n) $ has to equal $ b $ but I don't see how.

$\endgroup$
6
  • 2
    $\begingroup$ Just a hint: If $E(X_n) \to b$ in $L^2$, then we can have $\int |X_n - b| \ dP \leq \ldots$ Use Cauchy-Schwarz! $\endgroup$ Dec 25, 2019 at 5:12
  • $\begingroup$ Also: your last line only holds if all limits exist and are finite. We don't know if $E(X_n)$ exists! $\endgroup$ Dec 25, 2019 at 5:27
  • $\begingroup$ Ah I see. I have to check but I think we can assume the limits exist (as in it's in the problem statement). $\endgroup$
    – an1lam
    Dec 25, 2019 at 17:06
  • $\begingroup$ Maybe I'm just being dense but I don't see how the Cauchy-Schwarz inequality helps with the inequality chain you started. The probability version of Cauchy-Schwarz that I'm familiar with is $ \mathbb{E}(X^2Y^2) \leq \mathbb{E}(X^2) \mathbb{E}(Y^2) $. How does that relate to $ \mathbb{E} \lvert X_n - b \rvert \leq \dots $? $\endgroup$
    – an1lam
    Dec 26, 2019 at 20:49
  • 1
    $\begingroup$ Precisely. I'll do a full solution when I can figure the second part. $\endgroup$ Dec 29, 2019 at 23:31

2 Answers 2

4
+50
$\begingroup$

By Jensen's Inequality, (alternatively, this follows from noting $\operatorname{Var}(X_n - b) \geq 0$ ),$$\mathbb{E}(X_n - b)^2 \geq (\mathbb{E}|X_n - b|)^2$$ so taking the limit as $n\to\infty$ of both sides gives $0 \geq \limsup_{n\to\infty} \mathbb{E} |X_n - b|$, and we also clearly have $\liminf_{n\to\infty} \mathbb{E} |X_n - b| \geq 0$ since the argument is nonnegative. Then $\lim_{n\to\infty} \mathbb{E} |X_n - b| = 0$, so $\lim_{n\to\infty} \mathbb{E}(X_n) = b.$

For the second part, use the lemma posted from this stackexchange post. In particular, since $b$ is a constant, it has $0$ variance, so $\lim_{n\to\infty}\operatorname{Var}(X_n) = 0$.

$\endgroup$
0
0
$\begingroup$

The second part:

By the Var definition:

$$Var(X) = E(X^2) - [E(X)]^2$$

$$Var(X_n-b) = E(|X_n-b|^2) - [E(X_n-b)]^2$$

$$\lim_{n \to \infty}⁡Var(X_n-b) = \lim_{n \to \infty}⁡E(|X_n-b|^2 ) - \lim_{n\to \infty}⁡{[E(X_n-b)]^2}$$

the Quadratic mean tell us that:

$$ \lim_{n\to \infty} ⁡E(|X_n-b|^2 ) = 0$$

and the first part of this problem defined:

$$\lim_{n \to \infty}⁡(E|X_n-b|) = 0,$$

then

$$\lim_{n \to \infty}⁡ Var(X_n-b) = 0.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.