Finding expected value of remaining piece

Question

A father has a pie made for his two sons. Eating more than half of the pie will give indigestion to anyone. While he is away, the older son helps himself to a piece of the pie. The younger son then comes and has a piece of what is left by the brother. Assume that the size of each of the two pieces eaten by the sons is random and uniformly distributed over what is currently available.

What is the expected size of the remaining piece give that no son has indigestion?

After a while of thinking this is what I did. Let $\theta_1$ be the total angle corresponding to the amount of pie that son $1$ eats. Similarly, let $\theta_2$ be the total angle corresponding to the amount of pie for son $2$.

Let $X$ be the remaining piece. So, $$ X = 2\pi - \theta_1 - \theta_2 \\ \mathbb{E}(X) = 2\pi - \mathbb{E}(\theta_1) - \mathbb{E}(\theta_2). $$ So after a while of thinking, given that no son has indigestion, it feels as if son 1 and son 2 are independent, since they eat less than $\pi$...

So $\theta_1 \sim \operatorname{Unif}(0,\pi)$ and also $\theta_2\sim \operatorname{Unif}(0,\pi)$. Then $\mathbb{E}(X) = \pi$.

Is this correct?

Answer 1

$\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{} }$As is pointed out by @AndreaL, conditional probability should be used to calculate to desired probability.

Suppose the elder son eats an angle of $X_1$ and the younger son eats an angle of $X_2$. Translating to probabilistic notation, there is $X_1 \sim U(0, 2π)$ and $X_2 \mid X_1 \sim U(0, 2π - X_1)$, and what needs to be found is $E(2π - X_1 - X_2 \mid X_1 \leqslant π, X_2 \leqslant π)$.

The joint density of $(X_1, X_2)$ is$$ f_{X_1, X_2}(x_1, x_2) = f_{X_2 \mid X_1}(x_2 \mid x_1) f_{X_1}(x_1) = \frac{1}{2π(2π - x_1)}, \quad \forall x_1, x_2 > 0, x_1 + x_2 \leqslant 2π $$ thus\begin{align*} &\peq P(X_1 \leqslant π, X_2 \leqslant π)\\ &= \iint\limits_{0 < x_1, x_2 \leqslant π} \frac{1}{2π(2π - x_1)} \,\d x_1\d x_2 = \frac{1}{2π} \int_0^π \int_0^π \frac{1}{2π - x_1} \,\d x_2\d x_1\\ &= \frac{1}{2π} \int_0^π \frac{π}{2π - x_1} \,\d x_1 = \frac{\ln 2}{2}, \end{align*} and\begin{align*} &\peq E(2π - X_1 - X_2 \mid X_1 \leqslant π, X_2 \leqslant π)\\ &= 2π - E(X_1 + X_2 \mid X_1 \leqslant π, X_2 \leqslant π)\\ &= 2π - \iint\limits_{0 < x_1, x_2 \leqslant π} (x_1 + x_2)·\frac{f_{X_1, X_2}(x_1, x_2)}{P(X_1 \leqslant π, X_2 \leqslant π)} \,\d x_1\d x_2\\ &= 2π - \frac{1}{π\ln 2} \int_0^π \int_0^π \frac{x_1 + x_2}{2π - x_1} \,\d x_2 \d x_1\\ &= 2π - \frac{1}{π\ln 2} \int_0^π \frac{π x_1 + \frac{π^2}{2}}{2π - x_1} \,\d x_1\\ &= 2π - \frac{1}{π\ln 2}·\frac{1}{2} (5 \ln 2 - 2) π^2 = \frac{2 - \ln 2}{2\ln 2} π. \end{align*}

Answer 2

Let $x$ be the fraction of pie that the first son eats, and $y$ be the fraction of the remaining pie that the second one eats.

$x$ and $y$ are both taken uniformly in $[0;1]$.

Let $\theta_1 = x$ and $\theta_2 = y(1-x)$ be the fraction of the pie eaten by son 1 and 2 respectively.

You want to find out the expected value of $(1-x)(1-y)$ given that $\theta_1 \le \frac 12$ and $\theta_2 \le \frac 12$.

Let us try to find out the law of $(\theta_1,\theta_2)$ given those two inequalities.

for $z_1,z_2 \le \frac 12$,
$P_{z_1,z_2} = P(\theta_1 \le z_1 \land \theta_2 \le z_2) = P(x \le z_1 \land y(1-x) \le z_2)$ is the area of the corresponding shape inside the unit square, so it is :

$P_{z_1,z_2} = \int_{x=0}^{z_1}\int_{y=0}^{\min(1,z_2/(1-x))} dydx = \int_0^{z_1} \min(1,\frac {z_2}{1-x}) dx$

$1 < \frac {z_2}{1-x} \iff 1-x < z_2 \iff x > 1-z_2$.
Since $x \le z_1 \le \frac 12$ and $1-z_2 \ge \frac 12$, this is impossible, and so

$P_{z_1,z_2} = \int_0^{z_1} \frac {z_2}{1-x} dx = - \ln(1-z_1) z_2$.

Then, $P_{\frac 12,\frac 12} = \frac 12 \ln 2$, and so we finally have

$Q_{z_1,z_2} = P(\theta_i \le z_i \mid \theta_i \le \frac 12) = P_{z_1,z_2}/P_{\frac 12,\frac 12} = - \frac {\ln (1-z_1)}{\ln 2} (2 z_2)$.

Since $Q_{z_1,z_2} = f(z_1)g(z_2)$, you were surprisingly right when you said that $\theta_1$ and $\theta_2$ were independent. You were also right when you said that $\theta_2$ was uniformly distributed in $[0 ; \frac 12]$.
(those were your two statements where I immediately thought you were obviously wrong, which says something about how I am still dumb at probabilities)

However, $\theta_1$ (after conditioning) is not uniformly distributed in $[0 ; \frac 12]$ : if $\theta_1$ is small, there is a higher chance of the second son getting an indigestion than if $\theta_1$ is large. Conditioning on the second son not getting an indigestion then skews the distribution into a not uniform one.

Let us compute the expected fraction of pie eaten by the first son : Since $f'(t) = \frac 1{(1 - t)\ln 2}$,

this is $\int_0^{\frac 12} \frac t{(1 - t)\ln 2} dt = \frac 1 {\ln 2}\int_0^{\frac 12} (-1 + \frac 1{1-t})dt = \frac 1 {\ln 2}(\ln 2 - \frac 12) = 1 - \frac 1{2\ln 2} \approx 0.279 > \frac 14$.

And finally the expected fraction of remaining pie is $1 - (1 - \frac 1{2\ln 2}) - \frac 14 = \frac 1{2\ln 2} - \frac 14 \approx 0.471 < \frac 12$

Answer 3

This post only addresses (hopefully intuitively, without integration) WHY the two distributions are not independent $U(0, \frac{1}{2})$.

One can think of conditional probability as: try the experiment, then if the result doesnt fit the condition, ignore the result and retry the experiment; repeat until the result fits the condition.

Experiment A:

(Step A1) First son eats fraction $x$ of the pie, where $x \sim U(0, 1)$. If $x > \frac{1}{2}$, ignore the result, and retry (Step A1).

(Step A2) At this point we have $x < \frac{1}{2}$. Now second son eats fraction $y$ of the pie, where $y \sim U(0, 1-x)$. If $y > \frac{1}{2}$, ignore the result and retry (Step A2).

In experiment A, it is true that the final $x \sim U(0, \frac{1}{2})$, and the final $y$ also $\sim U(0, \frac{1}{2})$, and $x$ & $y$ are independent. HOWEVER this is not what the OP problem statement says! The OP says conditioned on both $x < \frac{1}{2}$ and $y < \frac{1}{2}$, so the correct experiment is:

Experiment B:

(Step B1) First son eats fraction $x$ of the pie, where $x \sim U(0,1)$.

(Step B2) Second son eats fraction $y$ of the pie, where $y \sim U(0,1-x)$.

(Step B3) If either $x > \frac{1}{2}$ or $y > \frac{1}{2}$, then retry from (Step B1).

The two experiments are NOT equivalent, and only B is the correct interpretation of the wording of the OP.

In particular, in experiment B, the final $x$ does NOT $\sim U(0, \frac{1}{2})$. This can be intuitively seen as follows. If the final $x$ is slightly less than $\frac{1}{2}$ after (Step B1), then chance of $y > \frac{1}{2}$ is $\approx 0$ in (Step B2) and so the chance the whole result needs to be discarded in (B3) is $\approx 0$. Meanwhile, if the final $x$ is almost $0$ after (B1), then the chance of $y > \frac{1}{2}$ is $\approx \frac{1}{2}$ in (B2) and so the chance the whole result needs to be discarded in (B3) is $\approx \frac{1}{2}$. Basically, after (B1), a larger $x$ has a SMALLER chance of being discarded in (B3). So the final $x$ has a better chance to be large (close to $\frac{1}{2}$) than to be small (close to $0$), and E[final $x$] strictly $> \frac{1}{4}$, so clearly final $x$ does not $\sim U(0, \frac{1}{2})$.

Answer 4

This is a complicated situation, but the question happens to greatly simply the process of answering it. The question states that no son has indigestion.

Therefore, neither son ate half the pie or more. (amount either son ate: $0 \le eaten \lt .5$)

Therefore, there is always some pie left. (amount left: $0 \lt remaining \le 1$)

Since the portions both sons ate are uniformly random and independent (since neither son eats more than half), in the expected case we just say that they both ate the middle value (halfway between 0 and .5, or .25). And since two sons ate, .25 + .25 = .5, half the pie is gone.

Another way to look at it is to see that the amount of pie left is uniformly random, and deduce that it is .5.

So while it is true that in the situation the amount the second son eats is highly dependent on the amount the first son eats, the amount they eat is independent for this specific question only.

So yes, you are correct.

Answer 5

In each stage, a son takes away a share of what is left and leaves the rest. The share is uniformly distributed over $[0,1]$ and hence the left share is also distributed uniformly over $[0,1]$. Let the whole cake be 1, after the first son has taken away some, it becomes $x$ and after the second son it will be $xy$ with both $x$ and $y$ uniformly distributed over $[0,1]$. $$\langle xy\rangle=\frac{\int_{\text{no indigestion}}xy\hspace{2mm}dxdy}{\int_{\text{no indigestion}}dxdy}$$ The "no indigestion" condition is $$x>\frac{1}{2}\hspace{3mm}\text{and}\hspace{3mm} x(1-y)<\frac{1}{2}$$ And the numerator becomes $$\int_\frac{1}{2}^1dx\int_{1-\frac{1}{2x}}^1dy\hspace{2mm}xy=\int_\frac{1}{2}^1dx\hspace{2mm}\frac{x}{2}\Big[1-\big(1-\frac{1}{2x}\big)^2\Big]$$ $$=\frac{1}{2}\int_\frac{1}{2}^1dx\hspace{2mm}\big(1-\frac{1}{4x}\big)=\frac{1}{4}\big(1-\frac{\log2}{2}\big)$$ The denominator is $$\int_\frac{1}{2}^1\frac{dx}{2x}=\frac{\log2}{2}$$ And the expected value $$\frac{1}{4}\big(\frac{2}{\log2}-1\big)\approx0.471$$