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Let $x$ be an indicator variable such that $E[X] = \frac{1}{3}$, calculate $E[1-x^{0.5}]$.

I'm having a hard time figuring out why this isn't equivalent to $E[1] - E[x^{0.5}]$ which would be $1-(\frac{1}{3})^{0.5}$

any help would be greatly appreciated.

Alt
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user3358192
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  • In general, it is not true that $E[g(X)] = g(E[X])$ and so $E[\sqrt{X}] \neq \sqrt{E[X]}$ as you are trying to use. Notice instead that an indicator variable takes on values $0$ and $1$ only, and for each of the two value of $X$, the value of $\sqrt{X}$ is $\ldots$ – Dilip Sarwate Mar 05 '14 at 02:36

2 Answers2

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I'm not sure what you mean by an indicator variable. If you mean a binary (0/1) variable then the answer is that $\sqrt{x}=x$ and hence $E[1-\sqrt{X}]= 1- EX=2/3$.

More generally, please note that $E[X^a]\neq (E[X])^a$ except in very special cases like when $a=1$ or $X$ is constant.

JPi
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  • The last statement is not true. In general, it only holds for $a=1$ , or $X$ is constant. No one ever says that $ E[X^2] = E[X]^2$. If so, then variance is always 0 (and which is why we need X to be constant). – Calvin Lin Mar 05 '14 at 02:37
  • Correct; too late. – JPi Mar 05 '14 at 02:47
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Hint: Because $X$ is an indicator variable, what is $ P (X = 0)$?

Hint: The linearity of expectation applies to the sum and product, but not to the composition of functions. In general,

$$ E[X^a] \neq E[X]^a.$$

Hint: Because $X$ is an indicator variable, hence $ E[X] = E[X^{0.5}]$.

Calvin Lin
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