How to determine the probability distribution of a variable obtained as function of another variable having its own probability distribution function?

Question

I have a variable, $y$, having the following probability distribution function:

$P_y(y)=\frac{|y-a+2ab|}{a^2b^2}\exp(-\frac{y-a+2ab}{ab})$ defined for $x\in\mathbb{R};x>=a(1-2b)$

where $a$ and $b$ are parameters.

I have another variable, $E$, which depends on $y$:

$E(y)=ce^{-\alpha(y-\beta)}$ defined for $x\in\mathbb{R};x>=a(1-2b)$

where $\alpha,\beta$ and $c$ are parameters.

I need to determine the probability distribution function for $E,\;P_E(E)$.

I had a try and I am going to explain.

The probability that $E\in[E_1,E_2]$ is given by the following integral:

$\int\limits_{E_1}^{E_2}{P_E(E)}dE$

and, (for an injective and continuous function $E(y)$, such as mine), is equal to the probability that $y\in[y_1,y_2]$, with $E_1=E(y_1)$ and $E_2=E(y_2)$, that is the integral:

$\int\limits_{y_1}^{y_2}{P_y(y)}dy$.

So we can write:

$\int\limits_{E_1}^{E_2}{P_E(E)}dE=\int\limits_{y_1}^{y_2}{P_y(y)}dy$

Now, going to infinitesimals:

$[E_1,E_2]\rightarrow[E',E'+dE']$ and $[y_1,y_2]\rightarrow[y',y'+dy']$

so that

$\int\limits_{E(y')}^{E(y'+dy')}{P_E(E)}dE=\int\limits_{E'}^{E'+dE'}{P_E(E)}dE=\int\limits_{y'}^{y'+dy'}{P_y(y)}dy$

but the last two integrals are evaluated over an infinitesimal interval, so that:

$P_E(E)dE=P_y(y)dy\Leftrightarrow P_E(E)=P_y(y)\left(\frac{dE}{dy}\right)^{-1}$.

That's it, but I don't think it is correct... What it the correct reasoning?

Thank you in advance

Answer 1

You have a probability density function (pdf) for the random variable $Y$:

$$g(y)=\frac{(2 a b-a+y) \exp \left(-\frac{2 a b-a+y}{a b}\right)}{a^2 b^2}$$

with the condition that $a b>0$.

The cumulative distribution function (cdf) of $Y$ being

$$\text{Pr}(Y \leq y)=\int_{a (1-2 b)}^y g(x) \, dx=\frac{e^{\frac{a-y}{a b}-2} (-3 a b+a-y)}{a b}+1$$

You have a random variable $E$ which is a function of $Y$:

$$E=c \exp (-\alpha (y-\beta ))$$

The cdf for $E$ is

$$\text{Pr}(E\leq e)=\text{Pr}(c \exp (-\alpha (Y-\beta ))\leq e)=\text{Pr}\left(Y \geq \beta -\frac{\log \left(\frac{e}{c}\right)}{\alpha }\right)=1-\text{Pr}\left(Y \leq \beta -\frac{\log \left(\frac{e}{c}\right)}{\alpha }\right)=\frac{\text{exp}\left({\frac{a-\beta +\frac{\log \left(\frac{e}{c}\right)}{\alpha }}{a b}-2}\right) \left(3 a b-a+\beta -\frac{\log \left(\frac{e}{c}\right)}{\alpha }\right)}{a b}$$

This results in the pdf of $E$ being

$$\frac{\text{exp}\left({\frac{a-\beta }{a b}-2}\right) \left(\frac{e}{c}\right)^{\frac{1}{a \alpha b}} \left(\alpha (a (2 b-1)+\beta )-\log \left(\frac{e}{c}\right)\right)}{a^2 \alpha ^2 b^2 e}$$

for $0\leq e \leq c \text{ exp}^{-\alpha (a (1-2 b)-\beta )}$.