1
$\begingroup$

Suppose that $X_1 , . . . , X_n$ form a random sample of size $n$ from the uniform distribution on the interval [0, 1]. Define a new random variable $Y_n=\max\{X_1,..,X_n\}$.

I know that i can compute the c.d.f. of $Y_n$ as follows $$P(Y_n<y)=P(\max\{X_1,..,X_n\}<y)=\prod_{j=1}^n P(X_j<y)= [P(X_j<y)]^n$$

I saw in my book that the c.d.f. of $Y_1$ can be written as $P(Y_1<y)=[1-P(X_j<y)]^n$. But why? I know that the $\max\{\}$ and $\min\{\}$ applications are kind of inverse one each other. But in my mind $Y_n$ and $Y_1$ have the same distribution.

Why $P(Y_1<y)$ is not like this $P(Y_1<y)=P(\min\{X_1,..,X_n\}<y)=\prod_{j=1}^n P(X_j<y)$ ?

I don't realize the consequence of having $\max\{\}$ or $\min\{\}$.

$\endgroup$
  • $\begingroup$ Yes. I did it because the $X_k$'s are taken at random. $\endgroup$ – Gabriel Sandoval Nov 1 '17 at 2:10
  • 1
    $\begingroup$ As for why $Y_{(n)}$ and $Y_{(1)}$ don't have the same distribution: Consider, as an example, drawing $100$ copies of a uniform random variable on $[0, 1]$. Note that $Y_{(n)}$ should be rather close to $1$ and $Y_{(1)}$ should be rather close to $0$. In particular, it is exceedingly unlikely that $Y_{(n)}$ will be less than $0.5$ or that $Y_{(1)}$ will be more than $0.5$. $\endgroup$ – Aaron Montgomery Nov 1 '17 at 2:44
0
$\begingroup$

If $X_1, X_2, \ldots, X_n$ is your sample, and $Y_1 = \min\{X_1, X_2, \ldots, X_n\}$, then $$\Pr[Y_1 \le y] \ne \prod_{j=1}^n \Pr[X_j \le y],$$ because the right hand side requires all of the $X_j$ to be less than or equal to $y$, whereas the left hand side only requires the smallest of the $X_j$ to be less than or equal to $y$. Thus the left hand side counts events that the right hand side does not. For example, if $n = 5$ and we have $(X_1, X_2, X_3, X_4, X_5) = (0.5, 0.3, 0.2, 0.8, 0.4)$, then $Y_1 = 0.2$. But the event $Y_1 \le 0.6$ is true, whereas the event $X_4 \le 0.6$ is false.

So this might make you ask why such an argument does work for $Y_n = \max\{X_1, X_2, \ldots, X_n\}$? Well, the event $Y_n \le y$ requires the largest of the $X_j$ to be less than or equal to $y$, which in turn implies that all of the $X_j$ are less than or equal to $y$, owing to the fact that $Y_n$ is by construction the largest member of the sample. This is in contrast to the same reasoning applied to $Y_1$: simply saying that the smallest member of the sample is less than or equal to some given number $y$, does not guarantee that the other members of that sample are also less than or equal to $y$. And that's the difference.

How would we modify the successful reasoning for computing $\Pr[Y_n \le y]$ to compute $\Pr[Y_1 \le y]$? The answer is to understand that the natural ordering of the minimum member as being the smallest member of the sample--thus every other member is at least as large--means we should consider the complementary probabilities: $$\Pr[Y_1 > y] = \prod_{j=1}^n \Pr[X_j > y].$$ And if we want to write things out in terms of cumulative distribution functions, we would just say $$\Pr[Y_1 \le y] = 1 - \Pr[Y_1 > y] = 1 - \prod_{j=1}^n \Pr[X_j > y] = 1 - \prod_{j=1}^n (1 - \Pr[X_j \le y]).$$

$\endgroup$
2
$\begingroup$

Having $\max\{X_1, \dots, X_n\} \leq y$ is equivalent to having $X_1 \leq y$ and $X_2 \leq y$ and $\dots$ and $X_n \leq y$. This is why it's convenient to work with in terms of distributions; the fact that we can express this as several independent events separated by the word "and" lets us multiply. But, the same can't be said for the minimum.

What we could say for the minimum is: having $\min\{X_1, \dots, X_n\} \leq y$ is equivalent to have $X_1 \leq y$ or $X_2 \leq y$ or $\dots$ or $X_n \leq y$. However, the word "or" is not nearly as convenient as the word "and" when it comes to probabilities. So, we use a trick instead; we observe that having $\min\{X_1, \dots, X_n\} \geq y$ is equivalent to having $X_1 \geq y$ and $X_2 \geq y$ and $\dots$ and $X_n \geq y$. This is why there are $(1-)$ terms both inside and outside the CDF expressions you presented; we're dealing with the opposite thing typically encoded by a CDF.

$\endgroup$
  • $\begingroup$ As Jack noted in the comments to the original post, I implicitly assumed that your random variables were independent. Based on the formulas you presented, that should be the case. $\endgroup$ – Aaron Montgomery Nov 1 '17 at 2:05
2
$\begingroup$

The event for the maximum of a sequence of random variables being less than or equal to a value, is the event for all of the variables to be so.   When the variables are independent, the probability for the conjoint event is the product of the probabilities of the events.   When the events are identically distributed, the probabilities are identical.

$$\begin{align}\mathsf P(Y_n\leqslant y) {}&=\mathsf P(\max{\{X_i\}}_{i=1}^n\leqslant y) \\[1ex]&= \mathsf P(\bigcap_{i=1}^n\{X_i\leqslant y\}) \\[1ex] &= \prod_{i=1}^n\mathsf P\{X_i\leqslant k\} \\[2ex] \mathsf P\{Y_n\leqslant y\} {}&= \mathsf P\{X_1\leqslant k\}^n\\[1ex] &= y^n~\mathbf 1_{y\in[0;1]}\end{align}$$


The event for the minimum of a sequence of random variables being greater than a value, is the event for all of the variables to be so.   Everything else follows as above, except that we are dealing with the complements of the events whose probability we seek and know.

$$\begin{align}\mathsf P\{Y_1\leqslant y\} {}&=\mathsf P(\min{\{X_i\}}_{i=1}^n\leqslant y) \\[1ex]&=1 -\mathsf P(\min\{X_i\}_{i=1}^n> y) \\[1ex] & = 1-\mathsf P(\bigcap_{i=1}^n\{X_i>y\}) \\[1ex] & = 1-\prod_{i=1}^n\mathsf P\{X_i>y\} \\[1ex] &= 1-\prod_{i=1}^n(1-\mathsf P\{X_i\leqslant y\}) \\[2ex]\mathsf P\{Y_1\leqslant y\} {} & = 1-{\big(1-\mathsf P(X_1\leqslant y)\big)}^n\\[1ex] &= (1-(1-y)^n)~\mathbf 1_{y\in[0;1]}\end{align}$$


Notice: This is not quite what you claim the book says, though it is likely what the book actually said.


But in my mind $Y_n$ and $Y_1$ have the same distribution.

Clearly the value for $Y_n$ will certainly be at least as great as that for $Y_1$.   Since $Y_n\geqslant Y_1$ almost surely, therefore they cannot follow the same distribution.   The least value in the sample is more likely than the greatest, to be smaller than some contant. $$\mathsf P(Y_1{\leqslant}y)\geq\mathsf P(Y_n{\leqslant} y)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.