4.2.2 Solution

We first note the following known probabilities in this problem. The probability of picking box \(S\) or box \(\bar {S}\) is \(\frac {1}{2}\). Once we pick box \(S\), then we have to switch the box. If we pick the \(\bar {S}\) box, then we switch only if the observed \(X\) is less than 1.

To help solve this problem, we start by drawing the decision tree describing the possible flow and assign a probability to each branch. At the end of each branch we draw the PDF of \(Y\) resulting from traversing that branch only. Next, we combine (add algebraically) all the PDF’s together after we scale each PDF by the probabilities found along the edges which lead to the end of the branch.

Using the above diagram as a guide, we now calculate the PDF for \(Y\) as follows (starting from the right most branch to the left most branch)

\begin{align*} P(Y & =y)=\frac {1}{2}\times \left \{ \begin {array} [c]{ccc}\frac {1}{2} & & 0<y<\frac {1}{2}\\ \frac {1}{2} & & \frac {1}{2}<y<1\\ \frac {1}{2} & & 1<y<2 \end {array} \right . +\frac {1}{2}\times \frac {1}{2}\times \left \{ \begin {array} [c]{ccc}2 & & 0<y<\frac {1}{2}\\ 0 & & \frac {1}{2}<y<1\\ 0 & & 1<y<2 \end {array} \right . +\frac {1}{2}\times \frac {1}{2}\times \left \{ \begin {array} [c]{ccc}0 & & 0<y<\frac {1}{2}\\ 0 & & \frac {1}{2}<y<1\\ 1 & & 1<y<2 \end {array} \right . \\ & \\ & =\left \{ \begin {array} [c]{ccc}\frac {1}{4} & & 0<y<\frac {1}{2}\\ \frac {1}{4} & & \frac {1}{2}<y<1\\ \frac {1}{4} & & 1<y<2 \end {array} \right . +\left \{ \begin {array} [c]{ccc}\frac {1}{2} & & 0<y<\frac {1}{2}\\ 0 & & \frac {1}{2}<y<1\\ 0 & & 1<y<2 \end {array} \right . +\left \{ \begin {array} [c]{ccc}0 & & 0<y<\frac {1}{2}\\ 0 & & \frac {1}{2}<y<1\\ \frac {1}{4} & & 1<y<2 \end {array} \right . \\ & \\ & =\left \{ \begin {array} [c]{ccc}\frac {3}{4} & & 0<y<\frac {1}{2}\\ \frac {1}{4} & & \frac {1}{2}<y<1\\ \frac {1}{2} & & 1<y<2 \end {array} \right . \end{align*}

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