---
title: "東京大学大学院　情報理工入試過去問（数学）解答・２０２２年　第三問"
tags: u-tokyo, exam
---

# 東京大学大学院　情報理工入試過去問（数学）<br><span style="float: right;">―解答・２０２２年　第三問</span>

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+ こちらの解答は非公式で個人記録用に作成されており、大学院側に認められたものではありません。正確性についての保証は致しかねます。
+ 問題本文は研究科のウェブサイトから見ることができます。
+ 満足出来た解答でしたらコメントしていただけると幸いです。

## ２０２２年　第三問
Let $A$ be a random point among the region $R: 0 \leq x \leq 1, 0 \leq y \leq 1$, having uniform distribution.

Let:
+ $\overline{AB}$ be the perpendicular line from $A$ to the $y$ axis,
+ $\overline{AC}$ be the perpendicular line from $A$ to the $x$ axis,
+ $O$ be the origin.
+ $S$ be the area of the rectangular area $ABOC$.

### (1) Find the expected value of $S$.
Let $X, Y$ be the independent random value of the $x, y$ coordinate of point $A$, both having uniform distribution.

Since they are uniformly distributed among $[0, 1]$, the probability density function is $f(x) = \frac{1 - 0}{1 - 0} = 1$.

Then,
$$
\begin{align}
\excp{S} = \excp{XY} = \excp{X}\excp{Y} &= \int^1_0xf_X(x)dx \cdot \int^1_0xf_Y(y)dy \\[1ex]
&= \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}
\end{align}
$$

### (2) Let there be some $r$ where $0 < r < 1$. Find the probability of $S < r$.
To let $S < r$, point $A$ must be in a reigon such that $XY < r$, which is the area $xy < r$ within $R$.

Since the point is randomly chosen, $\prob{A \leq r}$ is simply the area under $xy = r$, or $y = \frac{r}{x}$ within the range, divided by the area of the whole range, which is $1$.


to calculate the area, we divide into two pieces:
+ The rectangular area with point $M(r, 1)$ at the upper right, whose area is $1 \cdot r = r$

+ The rest of the area can be calculated as:
$$
\int^1_r \frac{r}{x} dx = \left. r(\ln x) \right\vert^{x = 1}_{x = r} = -r\ln r
$$

Thus the total area is $r - r\ln r$, inducing:
$$
\prob{S < r} = \frac{r - \ln r}{1 \cdot 1} = r - r\ln r
$$

### (3) Find probability density function of $S$.
$P(S < r)$ is the cumulative density function $\cdfsub{S}{r}$ of $S$. To obtain the probability density function, we can simply differentiate the cdf:
$$
\begin{align}
\frac{d}{dr} \cdfsub{S}{r} &= \frac{d}{dr} r - r\ln r \\[1ex]
&= \frac{d}{dr} r - \frac{d}{dr} r\ln r \\[1ex]
&= 1 - \left( \ln r \frac{d}{dr} r + r \frac{d}{dr} \ln r \right) \\[1ex]
&= 1 - (\ln r + 1) = -\ln r
\end{align}
$$

Thus, the probabiloty density function of $R : f_S(r) = -\ln r$.

### (4) Find the probaability density function $f_{_Z}(z)$of $Z$, defined by the following rules.
:::info
+ Let there be $n$ points $\nobj{A}{n}{,}$ decided independently randomly in region $R$
+ Let $S_i$ be the area defined by $A_i$.
+ Let $Z$ be the minimal among $\nobj{S}{n}{,}$
:::

For some $0 < z < 1$, the following statement are the same:
+ *minimum is $z$*
+ *at least one is less or equal to $z$*
+ **not** *all are greater greater than $z$*:
$$
\prob{Z \leq z} = \prob{\min(\nobj{S}{n}{,}) \leq z} =
1 - \prob{S_1 > z\;\cap\;S_2 > z\;\cap\; \ldots\;\cap\;S_n > z}
$$

Since
+ For each pair of $S_i, S_j, i \neq j$, $\prob{S_i \geq z}$ are independent with $\prob{S_j \geq z}$
+ All $S_i$ have the same probability density function
We can further say:
$$
\prob{Z < z} = 1 - \left( \prob{S_1 \geq z} \cdot \prob{S_2 \geq z} \cdot \ldots \cdot \prob{S_n \geq z} \right) = 1 - (1 - \prob{S < z})^n
$$

Substute the formula from $2$,
$$
\prob{Z < z} = 1 - (1 - (z - z\ln z))^n
$$

And to obtain the probability density function of $Z : f_{_Z}(z)$, we differentiate the cdf, getting:
$$
\begin{align}
f_{_Z}(z) = \frac{d}{dz}\cdfsub{Z}{z}
&= \cancel{\frac{d}{dz} 1} - \frac{d}{dz}\left( 1 - z + z\ln z \right)^n \\[1ex]
&= - \left[ \frac{d(1 - z + z\ln z)^n}{d(1 - z + z\ln z)} \cdot \frac{d(1 - z + z\ln z)}{dz} \right] \\[1ex]
&= -\left[ n(1 - z + z\ln z)^{n - 1} \cdot \ln z \right] \\[1ex]
&= -n(1 - z + z\ln z)^{n - 1}\ln z
\end{align}
$$