Expectation of a battery lifetime: Uniform Distributions

Question

Question:

A battery has a lifetime of $24$ hours and it is used for maximum three days. On each day, a person uses the battery for $K$ hours, where $K$ is uniform on $[0,24]$ and independent of the other days. What is the expectation of the remaining lifetime at the end of the third day?

My work:

Let $X$, $Y$, $Z$ as the amount of battery lifetime used on each day, respectively. Then the amount of battery lifetime left at the third day would be as follows:

$W=$ $[(24-X)/24]$ * $[(24-X-Y)/24]$ * $(24-X-Y-Z)$

Since at the second day, for a probability of $X/24$, the remaining lifetime would be $0$, and else (for a probability of $(24-X)/24$), there would be $(24-X-Y)$ left. I did the same for the third day.

Then I computed $E[W]$, which seemed to be $4$, but the correct answer seems to be $1$ (hour).

I think $W$ is miscalculated, because I checked the rest a couple times. Can anyone tell me what is wrong?

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(Edit) If it is simplified into a two-day process, I thought it can be calculated as the following:

Let $W$ the amount of lifetime left at the end of the second day. If $X$ is used on the first day, there is a $(X/24)$ probability that the battery will be used up on the second day. On the other hand, $((24-X)/24)$ is the probability that the used amount of the battery is $(X+Y)$, or $(24-X-Y)$ is the left amount.

Then $W=(X/24)*0+((24-X)/24)*(24-X-Y)$

and

$E[W]= (1/24) * E[(24-X)(24-X-Y)]= (1/24)*(24^2 - 48E[X] - 24E[Y] + E[X^2] + E[XY])$

where

$E[X]=E[Y], E[X^2]=E[XY]$ since $X$ and $Y$ are independent, and $E[X]=12, E[X^2]=192$.

Then we have:

$E[W]=24-2*12-12+16=4$ (hours),

which is the same as the calculation obtained by the integration method.

Answer 1

Don't you just want to integrate $24-x-y-z$ over the region where this is positive? That is, $${1\over24^3}\int_0^{24}\int_0^{24-x}\int_0^{24-x-y}(24-x-y-z)dzdydx$$

Answer 2

Just an addendum to saulspatz's answer, in which for convenience I'll change time units to days so every $24$ becomes a $1$: if the battery is used for uniformly distributed time periods on each of $n$ days, the fraction of its lifetime remaining is $\frac{1}{(n+1)!}$, which in the $n=3$ case considered herein becomes $\frac{1}{24}$, i.e. an hour as required. But where did I get this formula from?

We wish to evaluate $\int_{P_n}(1-\sum_{i=1}^n x_i)d^n x$, where $P_n$ is the region $x_i\ge 0,\,\sum_i x_i\le 1$. With the substitution $x_i=\sin^2 t_i\cdot \prod_{j<i}\cos^2 t_j$, the Jacobian is $2^n\prod_i \sin t_i \cos^{2n+1-2i}t_i$, and $1-\sum_i x_i=\prod_i\cos^2t_i$. The integral becomes $$2^n\prod_i\int_0^{\pi/2}\sin\theta\cos^{2n+3-2i}\theta d\theta=2^n\prod_i\frac{1}{2n+4-2i}=\prod_i\frac{1}{n+2-i}=\frac{1}{(n+1)!}.$$