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Let's say we have a random variable $X$ of which the distribution is unknown. Now there are these general rules like $E[X + Y] = E[X] + E[Y]$ etc. But what if we would define

$ \quad Y = \dfrac{1}{X} $

and we would be interested in the expected value $E[Y]$ and the variance $Var(Y) = E[(Y-\bar{Y})^2]$?

Now, I do realize that $X$ might be zero and therefore it is undefined. But what if we would assume $X \neq 0$? My quick solution was to assume $X$ being a log-normal random variable. But I would prefer something more general. Maybe a power series expansion is possible?

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Beyond $X\ne0$, you need $X$ to have zero density at $0$, otherwise $E[1/X]$ and $\operatorname{Var}(1/X)$ are both unbounded. –  Rahul Jan 2 '13 at 16:45

3 Answers 3

up vote 4 down vote accepted

If you write $X=\bar X+Z$, with $E[Z]=0$, then $$\begin{align} 1/X &= \sum_{k\geq0}(-1)^k\frac{Z^k}{\bar X^{k+1}} \\&= (1/\bar X)(1 - Z/\bar X + O(Z/\bar X)^2),\qquad (Z/\bar X \to0) \end{align}$$ and this series will converge by the ratio test if $|Z|<|\bar X|$ (in which case $X\neq0$ also). So then $$\begin{align}E[1/X]&=\sum_{k\geq0}(-1)^k\frac{E[Z^k]}{\bar X^{k+1}} \\&= \frac{1}{E[X]}+O\left(V[X]/\bar X^3\right),\qquad (Z/\bar X)\to0.\end{align}$$ The variance, in the limit $Z/\bar X\to0$, satisfies $$V[1/X]\to\frac{V[X]}{\bar X^4}.$$

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To put some mathematical rigor into these assertions, at present nothing more than heuristics, is a worthy enterprise. –  Did Jan 2 '13 at 17:12
    
Ah, now I see, you start with a Taylor expansion at $Z=0$ and then truncate it. I think that $|Z|<|\bar{X}|$ is indeed satisfied in my case. Thank you very much sir. –  demorge Jan 2 '13 at 17:53
    
What is O(*) in the expression above? Thanks. –  user90308 Aug 14 '13 at 9:42
    
@confusedstudent: en.wikipedia.org/wiki/Big_O_notation –  Willie Wong Aug 14 '13 at 13:23

Let's say that $X$ takes values in the interval $[a,b]$. (I realize that this is not the most general case, but this is an illustrative example.) We can then write

$$E[Y] = \int_a^b dx \: \frac{p(x)}{x} $$

where $p(x)$ is the PDF of $X$. The variance of $Y$ is then

$$ \mathrm{Var{Y}} = E[Y^2] - E[Y]^2 = \int_a^b dx \: \frac{p(x)}{x^2} - \left [ \int_a^b dx \: \frac{p(x)}{x} \right ]^2 $$

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I think it is fair to assume that OP knows what expectation and variance are, if he's talking about a log-normal random variable. –  Calvin Lin Jan 2 '13 at 16:40
    
Fair enough - I still thought it was helpful. –  Ron Gordon Jan 2 '13 at 16:42
    
Yes, I understand what you wrote, very nice. When I get home I'll have a look at where to introduce $p(x)$ in my problem. Thanks! –  demorge Jan 2 '13 at 17:01

If you restrict $X$ to be a positive random variable, you can apply Jensens to conclude that $E[\frac {1}{X}] \geq \frac {1}{E[X]}$. Equality holds if $X$ is a constant. There is no upper bound of $E[X]$.

There is no general formula. Given just $E[X], Var(X)$, you can construct probabilities such that $E[Y], Var(Y)$ take on any value.

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Well, not quite any value (as you showed yourself for E(Y)...). –  Did Jan 2 '13 at 17:10
    
@did Without the restriction of positive random variable, I believe there is nothing we can say about the general formula. I do not think that the asymptotics are useful (and I might be wrong about that). –  Calvin Lin Jan 2 '13 at 17:23

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