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In general can one say that for a random variable X:

$E[\frac{1}{X}] = \frac{1}{E[X]}$ ?

I've worked out a few examples where this works but I'm not sure how widely this is useful...

  • 1
    In general, $1/E[X]$ is the inverse of an arithmetic mean, and $E[1/X]$ is the inverse of a harmonic mean. Arithmetic and harmonic means of the same set are highly unlikely to be equal to each other...2012-12-01
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    If $X$ is a positive random variable, then this equality holds if and only if $X$ is a constant (that is, $X=c$ almost surely).2012-12-01
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    *I've worked out a few examples where this works*... Really? Which ones?2012-12-01

5 Answers 5

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It is very rarely true. Let's do a random example. Let $X$ be uniform on $[1,3]$. Then $E(X)=2$. But $$E\left(\frac{1}{X}\right)=\int_1^3 \frac{1}{x}\cdot\frac{1}{2}\,dx=\frac{\log 3}{2}\ne \frac{1}{2}.$$

For a simpler example, let $X=1$ with probability $1/2$, and let $X=3$ with probability $1/2$. Then $E(X)=2$.

But $E(1/X)=(1/2)(1)+(1/2)(1/3)=2/3$.

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For such a case, it is a good idea to study Jensen's inequality.

Another counterexample to the one given by André Nicolas is this one. Consider $X$ to be a normal distribution with mean $\mu$ and variance one. Then $E[X]=\mu$ but not only is $E[\frac{1}{X}]$ not in general equal to $1/\mu$; rather, it does not exist.

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    How do I see that it doesn't exist? Could you explain that? Thank you!2013-06-19
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    I think you'd need to compute the inner product of f(x) and 1/x (as per [wiki](https://en.wikipedia.org/wiki/Expected_value#General_definition)), which in this case (Wolfram Alpha notation ignoring constants, m=1: Integrate[exp(-(x-1)^2)/x,{x,-inf,inf}] ) just doesn't converge. Why? Well, I guess the area under the curve is just infinite. I remember [Infinite Acres](https://www.youtube.com/watch?v=V_c06ANw288) from calc, but beyond that you'd have to do some [tests](http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegralsCompTest.aspx) to really check.2016-03-11
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Jensen's inequality for functions of RVs is $\mathbf{E} \varphi(x) \geq \varphi(\mathbf{E}X)$ for convex functions and $\mathbf{E} \varphi(x) \leq \varphi(\mathbf{E}X)$ for concave functions. Clearly $Y = \frac{1}{X}$ is a convex function, so the first inequality holds.

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    There is a "for concave functions" missing.2013-07-02
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Y = 1/X is a convex function when x > 0

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    Alex already wrote that.2013-07-02
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In general if, X>0 , then the following inequality always will be satisfied:

E(1/X)>= 1/E(X)