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How to calculate the expectation of function of random variable without using probability density function? Note:- only cumulative distribution function is available. For example $E[g(X)]$=? where X is nonnegative r.v. with CDF $F_{X}(x)$.

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    $\begingroup$ If $X$ is a non-negative random variable, then $\int_0^{+\infty}(1-F_X(t))dt=E X$ (use Fubini's theorem). $\endgroup$ – Davide Giraudo Dec 16 '11 at 17:12
  • $\begingroup$ What do we know about $g$? Are there other assumptions than measurable? $\endgroup$ – Davide Giraudo Dec 16 '11 at 17:33
  • $\begingroup$ g(X) can be any function linear of exponential no further assumptions. $\endgroup$ – dikuve Dec 16 '11 at 17:43
  • $\begingroup$ Just for trivia, the rule you would use is called the "law of unconscious statistician", as you don't actually know the distribution of $g$. $\endgroup$ – r.g. Dec 16 '11 at 18:48
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If $X\geqslant0$ almost surely and if $g$ is regular, $$ \mathrm E(g(X))=g(0)+\int_0^{+\infty}g'(x)\cdot(1-F_X(x))\cdot\mathrm dx. $$ Proof: integrate with respect to $\mathrm P$ both sides of the almost sure relation $$ g(X)=g(0)+\int_0^{+\infty}g'(x)\cdot[x\lt X]\cdot\mathrm dx, $$ where $[\ \ ]$ denotes Iverson bracket.

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  • $\begingroup$ You omitted g'(x) in the second integral. I like it, it's a very nice generalisation of the equality mentioned by @DavideGiraudo in the first comment below the question. $\endgroup$ – savick01 Dec 16 '11 at 20:24
  • $\begingroup$ @savick01: Yes, thanks. $\endgroup$ – Did Dec 16 '11 at 20:27
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I'm going to hazard a guess about what the original poster might mean. Suppose wone knows the probability density function of $X$ but not of $g(X)$. My guess is that it is the latter that is not to be used. In that case, one can write $$ \mathbb{E}(g(X)) = \int_{-\infty}^\infty g(x)f(x)\;dx $$ where $f$ is the probability density function of $f$. In other words, one doesn't need to find the probability density function of $g(X)$.

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  • $\begingroup$ I am also not interested in finding the PDF of g(X). The question is How we can find the $E[g(X)]$ without using $f(x)$. Whereas we are having CDF of X i.e. $F_{X}(x)$. $\endgroup$ – dikuve Dec 17 '11 at 5:22
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At least formally, the probability density function is $f(X) = dF/dX$, so

$ E[g(X)] = \int dX f(X) g(X) = \int dX g(X) dF/dX = Fg(X_f) - Fg(X_i) - \int dX F(X) dg/dX $.

So if $g(X)$ is differentiable, and $g$ is finite at the endpoints of the domain of $X$ ($X_f$ and $X_i$), then you can try to evaluate the integral above.

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    $\begingroup$ Please: Don't use the capital $X$ to refer to two different things on the same line. First you wrote $E[g(X)]$, where $X$ is a random variable, then you used the same capital $X$ as the variable of integration. $\endgroup$ – Michael Hardy Dec 16 '11 at 19:18

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