How do I handle expectation in this problem?

Question

I recently failed to solve a problem relating to conditional expectation. The problem states:

A student is initially outside. He is given the choice with a probability of 0.4 of going to the playground, 0.3 of shopping in the mall and 0.3 of returning home. If he is in the playground, he takes 2 hrs of time and then decides to stay in the playground again or return back home. (he can continue to decide to stay in playground like a recursive process) Similarly, If he is in the mall, he takes 3 hrs of time and then decides to stay in the mall again or return back home. (he can continue to decide to stay in mall like a recursive process) For all cases, it takes 1 hr to return home. what is the expected hours the student was outside home?

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Initially, I thought about using Law of Total Expectation here. Like:

\begin{align*} E[X] &= E[X|Y=1]P(Y=1) + E[X|Y=2]P(Y=2) + \\ &\,\,\,E[X|Y=3]P(Y=3) \end{align*}

X = total # hrs outside, Y = choice.

But I don't know how to handle the conditional expectations for each individual event. For instance, in the case where students stay in the playground, it would be

$$E[X|Y=1] = 2 + 1/0.4.$$

Is it correct? Similarly for staying in the mall, it would be

$$E[X|Y=2] = 3 + 1/0.3.$$

Am I correct or am I making some very basic errors? Help would be appreciated.

Answer 1

Your formula for the law of total expectation is correct : $$\mathbb E[X] = \mathbb E[X\mid Y=1]\mathbb P(Y=1) + \mathbb E[X\mid Y=2]\mathbb P(Y=2) +\mathbb E[X\mid Y=3]\mathbb P(Y=3)\tag1 $$ here is how to deal with the conditional expectations:

• For the case $Y=1$ (when the student goes to the playground), you are being told that the student plays for 2 hours before deciding between playing 2 more hours or going home, in a recursive manner.
  By applying the law of total expectation to $X\mid Y=1$, we get the recursion $$\mathbb E[X\mid Y=1] = \mathbb E[X\mid Y=1, \text{student stays}] \mathbb P(\text{student stays}) + \mathbb E[X\mid Y=1, \text{student leaves}] \mathbb P(\text{student leaves})$$ Assuming that he stays in the playground with probability $0.4$, we get $$\mathbb E[X\mid Y=1] = (2+\mathbb E[X\mid Y=1])\cdot0.4 + 1\cdot0.6 $$ Which yields $\mathbb E[X\mid Y=1] \approx 2.33$
• By the same reasoning, we find for the case $Y=2$ that $$\mathbb E[X\mid Y=2] = (3 + \mathbb E [X\mid Y=2])\cdot0.3 + 1\cdot0.7 $$ From which you can deduce $\mathbb E[X\mid Y=2]$
• Lastly, the easiest case is $Y=3$, in which the student heads straight home (which takes one hour) : $$\mathbb E[X\mid Y=3] = 1 $$ You can now plug in all these values of the conditional expectations in the original equation $(1)$, and get an equation in $\mathbb E[X]$ that is easy to solve.