I have this question that I am not sure on how to deal with the $Y=cX$ part, can anyone help?

$X$ is a continuous random variable with $X$~$Exp(\lambda)$ and let $Y=cX$ where $c > 0$ is a fixed constant. I must find the probability density function (pdf) for $Y$ using two different methods:

1) Find the distribution function (cdf) for $Y$ and then differentiate it to find the pdf for $Y$.

I am fine with differentiation, I just am not sure how to find the cdf for $Y$

2) Noting that the function $g(x) = cx$ is strictly increasing and differentiable, find the pdf for $Y$ directly from the pdf for $X$

All I know is that you can write $X$~$\exp(\lambda)$ as:

$$f_X (x)=\left\{ \begin{array}{ll} 0 & \mbox{$x \le 0$};\\ \lambda e^{(-\lambda x)} & \mbox{$x < 0$}.\end{array} \right.$$

But I do not know what I can do with this (if anything).



CDF of random variable $Y$ is $P(Y \leq y)$, so we can write $P(Y \leq y) = P(cX \leq y) = P(X \leq \frac{y}{c})$

For the second method you can use $f_Y(y)=\frac{f_X(g^{-1}(y))}{g'(g^{-1}(y))}$

For more information about the formula this may be helpful: http://www.cs.unm.edu/~williams/cs530/gst2.pdf


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.