# Limits of Expectations

I've been fighting with this homework problem for a while now, and I can't quite see the light. The problem is as follows,

Assume random variable $X \ge 0$, but do NOT assume that $\mathbb{E}\left[\frac1{X}\right] < \infty$. Show that $$\lim_{y \to 0^+}\left(y \, \mathbb{E}\left[\frac{1}{X} ; X > y\right]\right) = 0$$

After some thinking, I've found that I can bound

$$\mathbb{E}[1/X;X>y] = \int_y^{\infty}\frac1{x}\mathrm dP(x) \le \int_y^{\infty}\frac1{y}\mathrm dP(x)$$

since $\frac1{y} = \sup\limits_{x \in (y, \infty)} \frac1{x}$ resulting in

$$\lim_{y \to 0^+} y \mathbb{E}[1/X; X>y] \le \lim_{y \to 0^+} y \int_y^{\infty}\frac1{y}\mathrm dP(x) = P[X>0]\le1$$

Of course, $1 \not= 0$. I'm not really sure how to proceed...

EDIT: $\mathbb{E}[1/X;X>y]$ is defined to be $\int_y^{\infty} \frac{1}{x}\mathrm dP(x)$. This is the notation used in Durret's Probability: Theory and Examples. It is NOT a conditional expectation, but rather a specifier of what set is being integrated over.

EDIT: Changed $\lim_{y \rightarrow 0^-}$ to $\lim_{y \rightarrow 0^+}$; this was a typo.

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It's not a good idea for titles to be entirely in $\TeX$. – J. M. Sep 6 '11 at 3:14
Srivatsan, J.M.: fixed! – duckworthd Sep 6 '11 at 3:16
@J.M. Interesting. Why is full title in TeX discouraged? – Srivatsan Sep 6 '11 at 3:30
To take myself as an example, @Sri, I browse questions by right-clicking on them to open in new tabs. Having the title entirely in $\TeX$ brings up the MathJax menu instead of the browser's popup menu, which interferes with this. Having a text portion allows you to have a place where right-clicking gives the browser's popup menu. – J. M. Sep 6 '11 at 3:48
@Sri yup, typo. Thanks for catching! – duckworthd Sep 6 '11 at 4:05

Hint(s) For every positive $y$, let $Z_y=(y/X)\mathbf{1}(y/X<1)$. You want to prove that $E(Z_y)\to0$ when $y\to0^+$.

It happens that $0\le Z_y<1$ with full probability, for every positive $y$, and that $Z_y\to0$ when $y\to0^+$ with full probability. (Of course you should check this.)

Now, your goal is to find a condition on a given family of nonnegative random variables $(T_y)$ that ensures that $$\lim\limits_{y\to0^+}E(T_y)=E\left(\lim\limits_{y\to0^+}T_y\right)$$ and to check that your family $(Z_y)$ fulfills this condition. There should not exist so many conditions of this ilk in your textbook...

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Got it! Thanks (again) @Didier! – duckworthd Sep 6 '11 at 7:37
@duck, you are welcome. – Did Sep 6 '11 at 8:52
@duckworthd Now that you solved the problem, could you mention what condition that is ? Thanks – Sasha Sep 6 '11 at 15:46
@Sasha, either bounded or dominated convergence should work here. – cardinal Sep 21 '11 at 23:13

Hint: For any $k > 1$, $\int_y^\infty \frac{y}{x} \ dP(x) \le \int_y^{ky}\ dP(x) + \int_{ky}^\infty \frac{1}{k} \ dP(x) \le \ldots$

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Hm... If I could split the limit into 2 ($\lim_{y \rightarrow 0^+} \int_{y}^{ky}dP(x) + \lim_{y \rightarrow 0^+} \int_{ky}^{\infty} \frac{1}{k} dP(x)$, one for both integrals, the first would go to 0 as $(y,ky) \rightarrow \emptyset$, but the second still wouldn't necessarily go to 0. I'm afraid I don't follow... – duckworthd Sep 6 '11 at 4:21
The second $\le 1/k$. – Robert Israel Sep 6 '11 at 4:43
Hmmm...My guess would be that you place a $\lim_{k \rightarrow \infty}$ outside of $\lim_{y \rightarrow 0^+}$, then let the first integral go to 0 as $y \rightarrow 0$ for fixed $k$, then swap the 2 $\lim$ and let the second integral go to 0 as $k \rightarrow \infty$ for fixed $y$. That doesn't even seem legal! – duckworthd Sep 6 '11 at 5:00
No, you don't swap any limits. For fixed $k$, when $y > 0$ is small enough your integral $< 2/k$. Therefore the limit must be $\le 0$. – Robert Israel Sep 6 '11 at 6:22
Another way is to make the split at $\sqrt{y}$ instead of $ky$. – Robert Israel Sep 6 '11 at 13:25