# expectation value 3

Suppose that $X$ is a non-negative random variable and there exist constants $A,B$ such that $$\forall t > 0\colon P(X>\frac{1}{t})<Bt$$ and $$\forall t > 0\colon E(\sin(tX))<At$$

I want to show that: $$E(X)<\infty$$

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I changed $\forall t\exists A,B$ to $\exists A,B\forall t$. Do you see why the question won't make sense otherwise? – Hagen von Eitzen Jan 15 '13 at 19:18

Here is a corrected version of Mihai Nica's argument. This was too long for a comment.

In the following, $E(\xi ; F)$ is a notation for $E(\xi\,1_F)$.

For every $n \geq 1$, we have $$E[\sin(X/n)\,;\, X \leq n] = E[\sin(X/n)] - E[\sin(X/n)\,;\,X > n]$$

Since $\sin(X/n) \geq -1$, we get $$E[\sin(X/n)\,;\, X \leq n] \leq E[\sin(X/n)] + P(X>n) \leq (A+B)/n$$

Using the fact that $n\sin(X/n)\,1_{\{X \leq n\}}$ converges almost surely to $X$ as $n \to \infty$ and stays non-negative, Pierre Fatou's lemma gives finally $$E(X) \leq \liminf_{n\to\infty}\;E[n\sin(X/n)\,;\,X \leq n] \leq A+B.$$

Note : we can get $E(X) \leq A + \dfrac{B}{\pi}$ in the same way.

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Notice, by linearity of $E$, that: $$E\left(\frac{\sin(tX)}{t}\right)<A$$ holds for every $t$. So its true then that $\liminf_{t\to 0} E\left(\frac{\sin(tX)}{t}\right)<A$. If only we could interchange the lim inf from outside the $E$ to inside, then we would be finished and happy, as $\liminf_{t\to 0} \frac{\sin(tX)}{t} = X$

This can be done using Fatou's Lemma, applied in a clever way. Define $X_n = \left.\frac{\sin(tX)}{t}\right|_{t=\frac{1}{n}}=n\sin{\frac{X}{n}}$ so that $\liminf_{n\to\infty} X_n = X$. Applying Fatou's lemma gives: (Actually there is one little hickup preventing us from applying Fatoe at this point that will be corrected later. Can you spot it?)

$$E(X)\leq \liminf_{n\to\infty}E(X_n) \leq A$$

And we are finished! There is one small problem though: Fatou's lemma does not apply because it is not true that $X_n$ is non-negative! Indeed $X_n$ is the sin of something so it can take both positive and negative values. To get around this obstacle, let us fix $M \in \mathbb{N}$ and restrict our attention to the set $\left\{ X < M \right\}$. On this set, we can be sure that $X_n \geq 0$ for $n > M$ as $X_n = n\sin{\frac{X}{n}}$ and the argument to the $\sin$ is $\leq 1$ and $\sin$ takes positive values here. We can now apply Fatou's lemma to the sequence $X_M, X_{M+1}, \ldots$ to get: $$E(X | X < M)\leq \liminf_{n\to\infty,n\geq M}E(X_n | X<M) \leq A$$ Since this holds for every $M$, and since the sets $\left\{X<M\right\} \uparrow \Omega$ increase to the whole probability space (Indeed, the first given inequality tells us $P(X>M) \to 0$ as $M\to\infty$), we can conclude $E(X) < A$

EDIT: The inequality $E(X_n|X<M) < A$ is wrong. To get around this, instead of fixing $M$, we should look at applying Fatou's lemma to $X_n 1_{X<n}$. See Ju'x's post for details.

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How do you justify that $E(X_n \mid X < M) \leq A$? – Siméon Jan 15 '13 at 23:14
Woops! That is an oversite on my part! I think its not true that $E(X_n | X< M) < A$. However I think we can fix it up using the first inequality (the one involving $B$). Namely: $E(X_n | X < M ) = E(X_n 1_{X<M}) / P(X<M) < E(X_n) / P(X<M) < A \frac{1}{1-B/M}$ For M large enough, this is $<2A$ – Mihai Nica Jan 15 '13 at 23:37
I think that the main problem lies in the inequality $E(X_n\,1_{X < M}) \leq E(X_n)$. – Siméon Jan 15 '13 at 23:42
Yes you are right. To fix it we should look at $E(X_n 1_{X <n})$ and use $E(X_n 1_{X <n}) < A + B$ as in your post. Thank you for pointing out the error! – Mihai Nica Jan 16 '13 at 2:37
@Mihai thanks for your answer – yalda Jan 16 '13 at 6:10