# Expectation of reciprocal of a variable

I am confused in applying expectation in denominator.

E(1/X)=?

can It be 1/E(X)?

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can it be 1/E(X)?

No, in general it can't; Jensen's inequality tells us that if $X$ is a random variable and $\varphi$ is a convex function, then $\varphi(\text{E}[X]) \leq \text{E}\left[\varphi(X)\right]$. If $X$ is strictly positive, then $1/X$ is convex, so $\text{E}[1/X]\geq 1/\text{E}[X]$, and for a strictly convex function, equality only occurs if $X$ has zero variance ... so in cases we tend to be interested in, the two are generally unequal.

I am confused in applying expectation in denominator.

$$\text{E}[g(X)] = \int_{-\infty}^\infty g(x) f_X(x) dx$$

(in the continuous case)

so when $g(X) = \frac{1}{X}$, $\text{E}[\frac{1}{X}]=\int_{-\infty}^\infty \frac{f(x)}{x} dx$

In some cases the expectation can be evaluated by inspection (e.g. with gamma random variables), or by deriving the distribution of the inverse, or by other means.

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As Glen_b say that's probably wrong, because the reciprocal is a non-linear function. If you want an approximation to $E(1/X)$ maybe you can use a Taylor expansion around $E(X)$:

$$E \bigg( \frac{1}{X} \bigg) \approx E\bigg( \frac{1}{E(X)} - \frac{1}{E(X)^2}(X-E(X)) + \frac{1}{E(X)^3}(X - E(X))^2 \bigg) = \\ = \frac{1}{E(X)} + \frac{1}{E(X)^3}Var(X)$$ so you just need mean and variance of X, and if the distribution of $X$ is symmetric this approximation can be very accurate.

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hello Matteo, is there any 2!, 3! etc missing in the denominator of the 2nd and 3rd terms???Because in Taylor's Series expansion all these terms will contain 2!, 3! etc in the denominators... –  Sandipan Karmakar May 5 at 9:31
Hi Sandipan, actually in the second order term the $2!$ in the denominator is cancelled by the fact that the second derivative of $1/x$ is $2 / x^3$. –  Matteo Fasiolo May 7 at 10:23
oh yes yes...I am very sorry that I could not apprehend that fact...I have one more q...Is this applicable to any kind of function???actually I am stuck with $|x|$...How can the expectation of $|x|$ can be deduced in terms of $E(x)$ and $V(x)$ –  Sandipan Karmakar May 8 at 11:58
I don't think you can use it for $|X|$ as that function is not differentiable. I would rather divide the problem into the cases and say $E(|X|) = E(X|X > 0)p(X>0) + E(-X|X < 0)p(X<0)$, I guess. –  Matteo Fasiolo May 8 at 20:35