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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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    Just for trivia, the rule you would use is called the "law of unconscious statistician", as you don't actually know the distribution of $g$.2011-12-16

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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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    @savick01: Yes, thanks.2011-12-16
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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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    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. $F_{X}(x)$.2011-12-17
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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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    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.2011-12-16