Let $X$ be a continuos r.v and X > 0 is this true? $\mathbb{E}(\frac{1}{X}) = \frac{1}{\mathbb{E}(X)}$
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1No... Try to come up with some counter-examples for yourself. – peter a g Jan 19 '21 at 01:59
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Does this answer your question? Is it true in general that $E(1/X) = 1/E(X)$? – 5201314 Jan 19 '21 at 02:07
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As an explicit example, take $X ( x)=x$ on the interval $[1,2]$, with uniform distribution $dx$ . – peter a g Jan 19 '21 at 03:40
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Does this answer your question? Expectation on 1/X – StubbornAtom Jan 19 '21 at 12:22
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Not in general:
$E\left(x\right)\:=\:\sum _{i=1}^n\:\left(x_i\cdot p_i\right)$
$E\left(\frac{1}{x}\right)\:=\:\sum _{i=1}^n\:\left(\frac{1}{x_i}\cdot p_i\right)$
Now, what happens when you plug numbers into that? Say X = {1,2,3,4} and p = {1/4, 1/4, 1/4, 1/4}. Then: E(X) = 2.5 whilst E(1/X) = 25/48.
However, that ws a mistake for omitting the key to the question: CONTINUOUS RV. Thus:
$E\left(x\right)=\int _{-\infty }^{\infty }\left(x\cdot f\left(x\right)\right)\:dx;\:E\left(\frac{1}{x}\right)=\int _{-\infty \:}^{\infty }\left(\frac{1}{x}\cdot \:f\left(x\right)\right)\:dx$
The intuition is retained from above. Most any other example will prove that E(x) doesn't equal E(1/x), except (maybe) for a some coincidences that may appear here and there.
Oscar
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