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I've worked on the following problem and have a solution (included below), but I would like to know if there are any other solutions to this problem, especially more elegant solutions that apply well known inequalities.

QUESTION: Suppose we have a random variable s.t. $P(a<X<b) =1$ where 0 < a < X < b , a and b both positive constants.

Show that $E(X)E(\frac{1}{X}) \le \frac{(a+b)^2}{4ab}$

Hint: find constant c and d s.t. $\frac{1}{x} \le cx+d$ when $a<x<b$, and argue that then we shall have $E(\frac{1}{X}) \le cE(X)+d$

MY SOLUTION: For a line $cx+d$ that cuts through $\frac{1}{X}$ at the points x=a and x = b, it's easy to show that $ c = - \frac{1}{ab} $ and $d = \frac{a+b}{ab} $,

So, $ \frac{1}{X} \le - \frac{1}{ab} X + \frac{a+b}{ab} $, and therefore:

$$ E(\frac{1}{X}) \le - \frac{1}{ab} E(X) + \frac{a+b}{ab} $$

$$ abE(\frac{1}{X}) + E(X) \le (a+b) $$

Now, because both sides of the inequality are positive, it follows that:

$$ [abE(\frac{1}{X}) + E(X)]^2 \le (a+b)^2 $$

$$ (ab)^2E(\frac{1}{X})^2 + 2abE(\frac{1}{X})E(X) + E(X)^2 \le (a+b)^2 $$

Then, for the LHS, we can see that $2abE(\frac{1}{X})E(X) \le (ab)^2E(\frac{1}{X})^2 + E(X)^2$

because

$0 \le (ab)^2E(\frac{1}{X})^2 - 2abE(\frac{1}{X})*E(X) + E(X)^2 = [abE(\frac{1}{X}) - E(X)]^2 $

SO,
$$ 4abE(\frac{1}{X})E(X) \le (ab)^2E(\frac{1}{X})^2 + 2abE(\frac{1}{X})E(X) + E(X)^2 \le (a+b)^2 $$

and therefore:

$$ E(\frac{1}{X})E(X) \le \frac{(a+b)^2}{4ab} $$ Q.E.D.

Thanks for any additional solutions you might be able to provide. Cheers!

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  • $\begingroup$ Did you try applying the hint? What happened? $\endgroup$ – Glen_b Mar 14 '15 at 23:43
  • $\begingroup$ I'd strongly suggest drawing a picture, and marking in a and b. A linear function with the property indicated in the hint should be obvious. $\endgroup$ – Glen_b Mar 15 '15 at 1:59
  • $\begingroup$ @Glen_b: Yes, I did try the hint, but as you can see above, the hint does not provide an obvious solution to the inequality (at least to one that I can readily see). $\endgroup$ – Lewkrr Mar 17 '15 at 0:35
  • $\begingroup$ Even though it's not something you might get credit for, as routine bookwork it still falls under self-study and follows the same rules. $\endgroup$ – Glen_b Mar 17 '15 at 0:39
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    $\begingroup$ Given your answer looks like quite a solid solution that actually uses the hint in what I believe is the intended fashion, I think you could post your solution as an answer. $\endgroup$ – Glen_b Mar 17 '15 at 4:01
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I know it's stated in the problem, but I figured I'd put it in the answer bank:

For some line $cx+d$ that cuts through $\frac{1}{X}$ at the points x=a and x = b, it's easy to show that $ c = - \frac{1}{ab} $ and $d = \frac{a+b}{ab} $,

So, $ \frac{1}{X} \le - \frac{1}{ab} X + \frac{a+b}{ab} $, and therefore:

$$ E(\frac{1}{X}) \le - \frac{1}{ab} E(X) + \frac{a+b}{ab} $$

$$ abE(\frac{1}{X}) + E(X) \le (a+b) $$

Now, because both sides of the inequality are positive, it follows that:

$$ [abE(\frac{1}{X}) + E(X)]^2 \le (a+b)^2 $$

$$ (ab)^2E(\frac{1}{X})^2 + 2abE(\frac{1}{X})E(X) + E(X)^2 \le (a+b)^2 $$

Then, for the LHS, we can see that $2abE(\frac{1}{X})E(X) \le (ab)^2E(\frac{1}{X})^2 + E(X)^2$

because

$0 \le (ab)^2E(\frac{1}{X})^2 - 2abE(\frac{1}{X})*E(X) + E(X)^2 = [abE(\frac{1}{X}) - E(X)]^2 $

SO,
$$ 4abE(\frac{1}{X})E(X) \le (ab)^2E(\frac{1}{X})^2 + 2abE(\frac{1}{X})E(X) + E(X)^2 \le (a+b)^2 $$

and therefore:

$$ E(\frac{1}{X})E(X) \le \frac{(a+b)^2}{4ab} $$

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    $\begingroup$ It's of interest that the RHS is the square of the ratio of the arithmetic mean to the geometric mean of the endpoints $(\text{AM}(a,b)/\text{GM}(a,b))^2$. While this upper bound comes from a secant of the $1/x$ function, it's also interesting to investigate the lower bound gained by drawing a tangent at the midpoint. $\endgroup$ – Glen_b Mar 18 '15 at 4:56
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Once we observe that both sides of the inequality are scale invariant, it follows immediately by combining two simple, well-known inequalities, of which the most notable is that correlation coefficients cannot be less than $\newcommand{\e}{\mathbb{E}}-1$.


The Cauchy-Schwarz Inequality guarantees that the correlation coefficient $\rho(X,Y)$ of any two random variables $X$ and $Y$ lies between $-1$ and $1$. Using the definition of correlation and focusing on the lower bound of $-1$ allows us to express this inequality in the form

$$-\text{sd}(X)\text{sd}(Y) \le \rho(X,Y)\text{sd}(X)\text{sd}(Y) = \text{Cov}(X,Y) = \e[XY] - \e[X]\e[Y].$$

If we let $Y=1/X$, the product $\e[X]\e[1/X]$ is recognizable right there at the very end (which of course is what inspired this approach).

Using the simplification $\e[XY]=\e[X/X]=\e[1]=1$, isolate the last term algebraically to obtain

$$\e[X]\e[1/X] \le 1 + \text{sd}(X)\text{sd}(1/X).$$

When a variable's values are confined to an interval $[a,b]$, its variance is limited by the value $(b-a)^2/4$. This is proven in several elegant, elementary, and informative ways at Variance of a bounded random variable; it comes down to the fact that variances cannot be negative. Consequently

$$\text{sd}(X)\text{sd}(1/X) \le (b-a)^2/4.$$

We may freely rescale $X$ because the left side of this inequality does not thereby change. Select a scale in which $ab=1$ (which is possible because both $a$ and $b$ are positive). This allows us to rewrite the preceding in a form where the right hand side is obviously scale invariant, too:

$$\text{sd}(X)\text{sd}(1/X) \le \frac{(b-a)^2}{4} = \frac{1}{ab}\frac{(b-a)^2}{4}=\frac{(b-a)^2}{4ab}.$$

These two inequalities finish the job:

$$\e[X]\e[1/X] \le 1 + \text{sd}(X)\text{sd}(1/X) \le 1 + \frac{(b-a)^2}{4ab} = \frac{(a+b)^2}{4ab}.$$

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