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Let $X$ be a random variable with an exponential distribution with $\lambda=1$ and $Y=2X$.

What is the density function of $f_y$?

I know that $$f_x =\begin{cases} e^{-x} & 0\leq x\leq\infty \\ 0 & \text{else} \end{cases},$$

but I'm not sure where to go from here. Would $$f_y =\begin{cases} 2e^{-x} & 0\leq x\leq\infty \\ 0 & \text{else} \end{cases}.$$

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Approach 1: find the CDF of $Y$ and differentiate.

The CDF of $Y$ is $$F_Y(y) = P(Y \le y) = P(2X \le y) = P(X \le y/2) = \cdots.$$ Taking the derivative with respect to $y$ yields the density.


Approach 2: change of variables. The derivative of the transformation $\phi(x) = 2x$ is $2$, so $$f_Y(y) = f_X(y/2) / 2.$$

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Find $F_Y(x) = \mathbb{P}[Y\leq x]$ and then differentiate it.

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