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The random variables $X_1,\ldots,X_n$ are independent and identically distributed with the cumulative distribution function $F$, which is absolutely continuous.

Let $Y=\max \lbrace X_i \rbrace$ and $Z=\min \lbrace X_i \rbrace$

How to compute the joint probability density function $g(a,b)$ of $Y$ and $Z$?

I used this:

I started to determine the cumulative distribution function of $Y$ and $Z$.

For $Y$ I got: $\mathbb{P}(Y \leq a)=\mathbb{P}(X_1 \leq a,\ldots, X_n \leq a) = \prod_{i=1}^n \mathbb{P}(X_i \leq a) =(F(a))^n$

For $Z$ I got: $\mathbb{P}(Z \leq b)=1-(1- \mathbb{P}(X_i \leq b))^n=1-(1-F(b))^n$

Since the joint cumulative distribution function is $G(a,b) = \mathbb P(Y \leq a, Z \leq b)=[F(a)]^n - [F(a) - F(b)]^n$, I want to compute the derivative to get $g(a,b)$:

$$g(a,b) = \frac{\partial^2}{\partial a \, \partial b} G(a,b),$$ so

$$ \frac{\partial}{\partial a} \frac{\partial}{\partial b} \left\{ [F(a)]^n - [F(a) - F(b)]^{n} \right\} = \frac{\partial}{\partial a} \left\{ -n[F(a) - F(b)]^{n-1} \cdot (-f(b)) \right\} \\ = \frac{\partial}{\partial a} \left\{ n[F(a) - F(b)]^{n-1} \cdot f(b) \right\} $$

So it's

$$\frac{\partial^2}{\partial a \, \partial b} G(a,b)=g(a,b)=\lbrace n(n-1)[F(a)-F(b)]^{n-2} \cdot f(b) \cdot f(a) \rbrace$$

Is this correct or is there something missing?

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  • $\begingroup$ I can only suggest, something like this. First note that whenever $x<y$ cdf is 0. Next note that $(\max {X_{i}}<x, \min {X_{i}}>y)$, is the same that $(X_{i}<x ; X_{i}>y)$. However you have $\max {X_{i}}<x, \min {X_{i}}<y$, But here you have $F_{Y,Z}(x,y)=F_{Z}(y)-F_{Y,Z}(x,y)$ . $\endgroup$
    – kolobokish
    Dec 14, 2019 at 14:50
  • $\begingroup$ Sorry, i made a mistake . $F_{Y,Z}(x,y)=F_{Y,Z}(x,\infty)-P(Y<x,Z>y)=F_{Y}(x)-P(Y<x,Z>y)$ $\endgroup$
    – kolobokish
    Dec 14, 2019 at 14:58
  • $\begingroup$ You've got it! Don't forget that this is only valid for $b \leq a$; if $b > a$, then the density is $0$. (Why?) $\endgroup$ Dec 15, 2019 at 18:51

1 Answer 1

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You'd like to be able to say something about the two-variable function $G(a,b) = \mathbb P(Y \leq a, Z \leq b)$ -- however, it's somewhat easier to deal with the quantity $\mathbb P(Y \leq a, Z \color{red}{\geq} b)$, because this is equal to $\mathbb P(b \leq X_i \leq a \text{ for all i}) = [F(a) - F(b)]^n$, where $F$ is the CDF of $X_i$. $\bigstar$

Since the variables are continuous, we have: $$\mathbb P(Y \leq a, Z \leq b) + \mathbb P(Y \leq a, Z \geq b) = \mathbb P( Y \leq a)$$ The middle term can be handled as above, and the term on the ride side is $[F(a)]^n$; thus, $$\mathbb P(Y \leq a, Z \leq b) = [F(a)]^n - [F(a) - F(b)]^n.$$

$\bigstar$ EDIT: I should note that this equality only holds when $b \leq a$; if $b > a$, then $\mathbb P(Y \leq a, Z \geq b)$ is just $0$.

SECOND EDIT: To get the joint density $g(a,b)$ from the joint distribution $G(a,b)$, use the relationship $$g(a,b) = \frac{\partial^2}{\partial a \, \partial b} G(a,b).$$

Although the relationship $f(x) = F'(x)$ is common, perhaps this one is not; note that

\begin{align*} \frac{\partial^2}{\partial a \, \partial b} G(a,b) &= \frac{\partial^2}{\partial a \, \partial b} \int_{-\infty}^a \int_{-\infty}^b f(x,y) \, \textrm d y \, \textrm d x \end{align*} and apply the fundamental theorem of calculus to the iterated integrals.


Applying this work to this problem: in the case where $b \leq a$, we would have \begin{align*} \frac{\partial}{\partial a} \frac{\partial}{\partial b} \left\{ [F(a)]^n - [F(a) - F(b)]^{n} \right\} &= \frac{\partial}{\partial a} \left\{ -n[F(a) - F(b)]^{n-1} \cdot (-f(b)) \right\} \\ &= \frac{\partial}{\partial a} \left\{ n[F(a) - F(b)]^{n-1} \cdot f(b) \right\} \end{align*} by the Chain Rule. Can you finish from here? (Don't forget to consider the case where $b > a$....)

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  • $\begingroup$ So $G_{Y,Z}(a,b)=\mathbb{P}(Y \leq a, Z \leq b)=[F(a)]^n-[F(a)-F(b)]^n=(\frac{a-x}{y-x})^n-((\frac{a-x}{y-x})-(\frac{b-y}{x-y}))^n$. Now I have trouble to calculate the derivation to get $h_{Y,Z}(a,b)$. I'm not sure if this works. $\endgroup$
    – Gerturter
    Dec 14, 2019 at 17:38
  • $\begingroup$ @Gerturter If you'd like to edit your question to show your trouble with the derivative, we could try to comment on it. I'm not sure how to help further given just this comment, though. (Please note that the variables in my formulation are $a, b$, so derivatives should be taken w.r.t. those.) $\endgroup$ Dec 14, 2019 at 17:50

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