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If $X$ is uniform on $[0, 1]$, and $Y$ is a discrete random variable which is either $1$ or $2$ with probability $\frac12$ each, find the cumulative distribution function of the product $XY$.

I know what the answer to the question is, I simply want to know HOW to solve this. The answer is: $$ F(t) = \begin{cases} 0 & \mbox{for } t < 0\\ \tfrac{3}{4}t & \mbox{for } 0 ≤ t < 1\\ \tfrac{t}{4} + \tfrac{1}{2} & \mbox{for } 1 ≤ t < 2\\ 1 & \mbox{for } t ≥ 2\end{cases} $$

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Find $P\{XY \leq t\}$ for a fixed number $t$ between $0$ and $1$. You will need to use the law of total probability to account for the cases that $Y=1$ and $Y=2$. Repeat for the case when $t$ is between $1$ and $2$ (in which case life is a little simpler because you know for sure that $Y$ necessarily has value $2$). – Dilip Sarwate Nov 9 '12 at 3:27

As Dilip had outlined, you the law of total probability: $$\begin{eqnarray} \mathbb{P}\left(X Y \leqslant t\right) &=& \mathbb{E}\left( \mathbb{P}\left(X Y \leqslant t | Y\right)\right) = \mathbb{E}\left( \mathbb{P}\left(X \leqslant \frac{t}{Y} | Y\right)\right) =\mathbb{E}\left( F_X\left(\frac{t}{Y}\right)\right) \\&=& F_X\left(t\right) \mathbb{P}\left(Y=1\right) + F_X\left(\frac{t}{2}\right) \mathbb{P}\left(Y=2\right) \\ &=& F_X(t) + F_X\left(\frac{t}{2}\right) \end{eqnarray} $$ It now remains to combine piecewise expressions for $F_X(t) = \max(0,\min(1,t))$.

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