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Given the random variable $Y = M\left( {1 - \frac{{1 + N}}{{{e^{a \times P \times X}} + N}}} \right)$ with the following real number $N>1, a>0, P>0, M>0$ and $X$ is an exponential random variable with the corresponding PDF: ${f_X}\left( x \right) = \frac{1}{\lambda }\exp \left( {\frac{{ - x}}{\lambda }} \right)$

I can find the CDF of Y as follow: $\Pr \left[ {Y < y} \right] = \Pr \left[ {M\left( {1 - \frac{{1 + N}}{{{e^{a \times P \times X}} + N}}} \right) < y} \right] = \Pr \left[ {{e^{a \times P \times X}} < \frac{{N + 1}}{{1 - \frac{y}{M}}} - N} \right] = \Pr \left[ {X < \frac{{\ln \left[ {\frac{{M + yN}}{{M - y}}} \right]}}{{a \times P}}} \right] = 1 - {\left( {\frac{{M + yN}}{{M - y}}} \right)^{\frac{{ - 1}}{{aP\lambda }}}}$

After that differentiate the CDF with respect to $y$, I obtain:

${f_Y}\left( y \right) = \frac{1}{{aP\lambda }}{\left( {\frac{{M + yN}}{{M - y}}} \right)^{ - 1 - \frac{1}{{aP\lambda }}}}\left( {\frac{N}{{M - y}} + \frac{{M + yN}}{{{{\left( {M - y} \right)}^2}}}} \right)$

How can I find the supporting set of $Y$ since I really want to know on which integration interval that the integration of $\int_{???}^{???} {{f_Y}\left( y \right)} = 1$?

Extra Information:

One thing that is very strange to me is that usually the PDF at infinity should be one however:

${F_Y}\left( \infty \right) = \mathop {\lim }\limits_{y \to \infty } \left[ {1 - {{\left( {\frac{{M + yN}}{{M - y}}} \right)}^{\frac{{ - 1}}{{aP\lambda }}}}} \right] = 1 - {\left( { - N} \right)^{\frac{{ - 1}}{{aP\lambda }}}}$

Tuong Nguyen Minh
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1 Answers1

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I also get $$F_Y(y)=1-\left(\frac{m+ny}{m-y}\right)^{-1/\lambda ap}=1-\left(\frac{m-y}{m+ny}\right)^{1/\lambda ap}$$

From the definition of $Y$ you can easily see, that $support(Y)=[0,m]$, by plugging in $X=0$ and $X\rightarrow\infty$. When you compute the corresponding limits, keep in mind, that $F_Y(y)$ is not defined on the whole real numbers, but only on an interval.

Naturally, one would extent $F_Y(y)$ to a function on the real numbers like this: $$\bar F_Y(y):=\begin{cases}0,&y<0\\F(y),&y\in[0,m]\\ 1,&y>m\end{cases}$$

Does this help?

Peter Strouvelle
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  • Thank you so much ! The integration $\int_0^M {{f_Y}\left( y \right)} = 1$ is now correct. However, I want to ask that does the method of $X=0$ and $X$ to Infinity was given based on the fact that the supporting set of $X$ was the half interval $\left[ {0,\infty } \right)$ ? – Tuong Nguyen Minh Jun 13 '21 at 15:18
  • Yes, the half-open interval $[0,\infty)$ is the support of X. – Peter Strouvelle Jun 13 '21 at 15:23