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$X$ is an exponential random variable with parameter $\frac{1}{\sigma^2}$.

If $Y=\frac{P_1X}{N_0}$ why is it exponentially distributed with parameter $\frac{N_0}{P_1\sigma^2}$?

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closed as off-topic by heropup, Clement C., user284331, NCh, JonMark Perry Apr 15 '18 at 2:37

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  • $\begingroup$ What are $P_1, N_0$? If they are just constants, then why write them instead of $Y=\alpha X$, where $\alpha>0$ is a constant (leading to a simpler, equivalent but more understandable question)? $\endgroup$ – Clement C. Apr 14 '18 at 21:09
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For questions where you transform variables, it is often useful to look at the CDF of the r.v. rather than the PDF.

If $X$ is exponentially distributed with parameter $1/\sigma^2$, then $$P(X\le x)=1-e^{-x/\sigma^2}$$Therefore we get$$P(Y\le y)=P\left(X\le \frac{N_0y}{P_1}\right)=1-\exp\left(-\frac{N_0 y}{P_1\sigma^2}\right)=1-\exp\left(-\frac{N_0}{P_1\sigma^2} y\right)$$This is the CDF for an exponential variable with parameter $\frac{N_0}{P_1\sigma^2}$. So $Y$ is distributed exponentially with this parameter.

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  • $\begingroup$ perfect thanks so much $\endgroup$ – Mokhtar Apr 14 '18 at 21:25

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