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Say we have an algorithm that takes time $T$ to process a problem of size $n$.

Is $\langle T(n)\rangle $ = $T(\langle n\rangle)$?

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  • $\begingroup$ What are your thoughts? What have you tried? $\endgroup$ – D.W. Oct 6 '16 at 17:16
  • $\begingroup$ I understand now in general it's not true that $\langle f(g) \rangle \neq f(\langle g \rangle)$. I guess that fact that $f$ here is an algorithm's runtime led me to confuse expectation w/ the mode - $T(n)_{mode} = T(n_{mode})$? But I do understand why $\langle f(g) \rangle \neq f(\langle g \rangle)$. I feel dumb. $\endgroup$ – Tushar Rakheja Oct 6 '16 at 21:36
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In general, it does not hold that $\DeclareMathOperator{\EE}{\mathbb{E}}\EE[f(X)] = f(\EE[X])$. For example, suppose that $X=0$ with probability $1/2$ and $X=1$ with probability $1/2$, and that $f(x) = x^2$. Then $\EE[f(X)] = 1/2$ whereas $f(\EE[X]) = 1/4$.

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