7

I've got a problem, with a solution, in my introduction to mathematical statistics book and I just don't get how they got there. There are follow up questions so I'd like to get insight at how they got to the answer.

The problem: Let $X_1,\ldots,X_n$ be independent random variables with the uniform distribution on the interval $[0,1]$. Determine the expectation and variance of $Y=\max(X_1,\ldots,X_n)$. Hint: Deduce the density of Y from the distriution function $P(Y \leq y)$ of $Y$, which can be determined using the distribution functions of $X_1, \ldots, X_n$.

The solution: $\operatorname E[Y]=\frac{n}{n+1}$, $\operatorname{var}[Y]=\frac{n}{n+2}+\left(\frac{n}{n+1}\right)^2$.

Now I know that a uniformly distributed random variable on $[0,1]$ has the following probability distribution, expectation and variance;

$F(x)=x$ for $x \in [0,1]$, $\operatorname E[X]=\frac{1}{2}$ and $\operatorname{var}[X]=\frac{1}{12}$.

saz
  • 123,507
Mathbeginner
  • 931
  • 4
    Have you tried to start with the hint? For $y\in[0,1]$, $$\mathbb{P}{ Y \leq y } = \mathbb{P}{ \max(X_1,\dots,X_n) \leq y } = \mathbb{P}{ \forall 1\leq i\leq n,\ X_i \leq y } = \mathbb{P}({ X_1 \leq y }\cap{ X_2 \leq y }\cap\dots\cap{ X_n \leq y }) = \dots$$ – Clement C. Sep 04 '18 at 19:55
  • Thanks for your comment. I would say all those individual probabilities are equal to $y$. Thus $ P({X_{1} \leq y} \cap ... \cap {X_{n} \leq y}) = y^n $. Can you confirm? – Mathbeginner Sep 04 '18 at 20:03
  • Incidentally, the solution for the variance is wrong: the "+" should be a "-". (Sanity check: when $n\to\infty$, the variance would converge to $2$.... you cannot have a variance greater than $1$ for a random variable in $[0,1]$!) – Clement C. Sep 04 '18 at 20:03
  • Yes, this is $y^n$ -- the reason is that the $X_i$'s are independent, so the probability of the intersection is equal to the product of the probabilities. Now, you have the cdf of $Y$. Use it to derive the expectation and variance. – Clement C. Sep 04 '18 at 20:04
  • One thing that may be useful (as a "trick" to remember): we have $$\mathbb{E}[Y] = \int_0^1 \mathbb{P}{Y\geq y} dy$$ and $$\mathbb{E}[Y^2] = \int_0^1 2y\mathbb{P}{Y\geq y} dy$$ See this – Clement C. Sep 04 '18 at 20:08
  • I expect the $\geq$ is a typo and should be $\leq$. But when I solve this for $E[Y]$ I get: $E[Y]=\int_0^1 y^n dy = \frac{1}{n+1} y^{n+1} \Big|_0^1=\frac{1}{n+1}1-\frac{1}{n+1}0=\frac{1}{n+1}$. Which doesn't equal the given solution. Where do I go wrong? – Mathbeginner Sep 04 '18 at 20:35
  • 1
    No, it is not a typo (see the link at the end of the comment). – Clement C. Sep 04 '18 at 20:37
  • https://math.stackexchange.com/q/751229/321264, https://math.stackexchange.com/q/745067/321264 – StubbornAtom May 10 '20 at 07:54

2 Answers2

10

Notice that $$\mathbb{P}(Y \leq y) = \mathbb{P}(\max(X_1, \dots, X_n) \leq y)\text{.}$$ Now, note that if the largest of $n$ numbers is less than $y$, it follows that each of the $n$ numbers must be less than $y$ as well.

So, $$\mathbb{P}(\max(X_1, \dots, X_n) \leq y) = \mathbb{P}(X_1 \leq y \cap X_2 \leq y \cap\cdots\cap X_n \leq y)\text{.}$$ By independence, we have $$\mathbb{P}(X_1 \leq y \cap X_2 \leq y \cap\cdots\cap X_n \leq y) =\mathbb{P}(X_1 \leq y) \mathbb{P}(X_2 \leq y)\cdots\mathbb{P}(X_n \leq y)=y^n$$ for $y \in [0, 1]$. Hence, $$F_{Y}(y)=\mathbb{P}(Y \leq y) = \begin{cases} 0, & y < 0 \\ y^n, & y \in [0, 1] \\ 1, & y > 1\text{.} \end{cases}$$ The PDF of $Y$ is thus $$f_{Y}(y)=F^{\prime}_{Y}(y) = \begin{cases} ny^{n-1}, & y \in [0, 1] \\ 0, & \text{otherwise.} \end{cases}$$

Thus, $Y$ follows a Beta distribution with $\alpha = n$ and $\beta = 1$. Hence, the mean is $$\mathbb{E}[Y]=\dfrac{\alpha}{\alpha+\beta}=\dfrac{n}{n+1}$$ and the variance is $$\text{Var}(Y) = \dfrac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)} = \dfrac{n}{(n+1)^2(n+2)}\text{.}$$

Clarinetist
  • 20,278
  • 10
  • 72
  • 137
9

Here is an alternative approach which doesn't require any knowledge of the Beta distribution.

Since

$$Y=\max\{X_1,\ldots,X_n\} \leq y \iff \forall j=1,\ldots,n: X_j \leq y$$

it follows from the independence of the random variables $(X_j)_{j \leq n}$ that

$$\mathbb{P}(Y \leq y) = \mathbb{P} \left( \bigcap_{j=1}^n \{X_j \leq y\} \right) = \prod_{j=1}^n \mathbb{P}(X_j \leq y) = y^n \tag{1}$$

for $y \in (0,1)$. Thus the distribution function $F$ of $Y$ satisfies

$$F(y) = \begin{cases} 0, & y \leq 0, \\ y^n, & y \in (0,1), \\ 1, & y \geq 1. \end{cases}$$

Approach 1: Differentiating $F$ we find that $Y$ has a density $p$ with respect to Lebesgue measure, $$p(y) = n y^{n-1} 1_{(0,1)}(y).$$ Thus $$\mathbb{E}(Y)= \int y p(y) \, dy =n \int_0^1 y^n \, dy = \frac{n}{n+1}$$ and $$\mathbb{E}(Y^2) = \int y^2 p(y) \, dy = n \int_0^1 y^{n+1} \, dy = \frac{n}{n+2}.$$ Thus, $$\begin{align*} \text{var}(Y) = \mathbb{E}(Y^2)-(\mathbb{E}(Y))^2 = \frac{n}{n+2} - \frac{n^2}{(n+1)^2} &= \frac{n(n+1)^2 - n^2 (n+2)}{(n+1)^2 (n+2)} \\ &= \frac{n}{(n+1)^2 (n+2)}. \end{align*}$$

Approach 2: For any non-negative random variable $Z$ it holds that

$$\mathbb{E}(Z) = \int_0^{\infty} \mathbb{P}(Z > z) \, dz = \int_0^{\infty} (1-\mathbb{P}(Z \leq z)) \, dz, \tag{2}$$

see e.g. this question for details. For $Z:=Y$ we obtain from $(1)$ that

$$\mathbb{E}(Y) = \int_0^{1} (1-y^n) \, dy= 1 - \frac{1}{n+1} = \frac{n}{n+1}.$$

Similarly,

$$\mathbb{E}(Y^2) \stackrel{(2)}{=} \int_0^{\infty} (1-\mathbb{P}(Y^2 \leq y)) \, dy = \int_0^{\infty} (1-\mathbb{P}(Y \leq \sqrt{y})) \, dy,$$

and so by $(1)$

$$\mathbb{E}(Y^2) = \int_0^1 (1- y^{n/2}) \, dy = 1- \frac{1}{\frac{n}{2}+1} = \frac{n}{n+2}.$$

As in the first approach, we thus find

$$\begin{align*} \text{var}(Y) = \mathbb{E}(Y^2)-(\mathbb{E}(Y))^2 = \frac{n}{n+2} - \frac{n^2}{(n+1)^2} = \frac{n}{(n+1)^2 (n+2)}. \end{align*}$$

saz
  • 123,507