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I am confused in applying expectation in denominator.


can It be 1/E(X)?

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2 Answers

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.

Use the law of the unconscious statistician

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

(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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