# What is the distribution of $Y=\frac{P_1X}{N_0}$ given $X\sim E(1/\sigma^2)$? [closed]

$X$ 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 $\frac{N_0}{P_1\sigma^2}$?

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

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• 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)? – Clement C. Apr 14 '18 at 21:09

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.