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Suppose $X$ is distributed exponentially with parameter $\lambda$. Its pdf is $\lambda e^{-\lambda x}$, and the calculation of its expectation is straight forward: $\mathbb{E}(X) = \int_0^\infty x\lambda e^{-\lambda x} dx=\lambda^{-1}$ (using integration by parts).

I now introduce this variant of the exponential distribution: let $H:[0,\infty)\to[0,\infty)$ be an increasing continuous bijection, and define $Y=H^{-1}(X)$. If I am not mistaken, the pdf of $Y$ is $\lambda e^{-\lambda H(x)}$. The idea behind the distribution of $Y$ is that the time, a term which is natural to the exponential distribution, moves not linearly.

Calculating the expectation of $Y$, in general, looks impossible. It is the result of $\int_{0}^\infty x\lambda e^{-\lambda H(x)}dx$. However, with some assumptions on $H$ I guess this can be solved.

The difficult question is this: can this integral be solved for $H(x)=\int_0^x h(t)dt$ such that $h$ is a rational function with $h(0)=0$?

The simpler question is this: can this integral be solved for $H(x)$ which is a polynomial?

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Actually the PDF of $Y$ is the function $g$ defined by $$ g(y)=\lambda h(y)\mathrm e^{-\lambda H(y)}, $$ where $h$ is the derivative of $H$, if $H$ is differentiable. In the general case, the CDF of $Y$ is the function $G$ defined by $$ G(y)=P(Y\leqslant y)=P(X\leqslant H(y))=1-\mathrm e^{-\lambda H(y)}. $$ In particular, $$ E(Y)=\int_0^\infty P(Y\gt y)\mathrm dy=\int_0^\infty\mathrm e^{-\lambda H(y)}\mathrm dy. $$

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  • $\begingroup$ Thanks! Do you mind explaining why the pdf is as you've written? $\endgroup$
    – Bach
    May 18, 2014 at 10:12
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    $\begingroup$ As usual, differentiate the CDF when the CDF is differentiable. $\endgroup$
    – Did
    May 18, 2014 at 10:13
  • $\begingroup$ Ah, sure, thanks! $\endgroup$
    – Bach
    May 18, 2014 at 10:24

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