# Prove symmetry of probabilities given random variables are iid and have continuous cdf

Let $Y_1, Y_2, ...$ be independent and identically distributed random variables in $(\Omega, \mathscr{F}, \mathbb{P})$ s.t. their distributions are continuous and $$F_{Y}(y) := F_{Y_1}(y) = F_{Y_2}(y) = ...$$

For $m = 1, 2, ...$ and $i \le m$, define $$A_{i,m} := (\max\{Y_1, Y_2, ..., Y_m\} = Y_i), B_m := A_{m,m}$$

Is it really that $$P(B_m) = P(A_{m-1,m}) = ... = P(A_{2,m}) = P(A_{1,m})$$

and hence $$P(B_m) = 1/m?$$

I guess $(A_{(i,m)})_{i \le m}$'s pairwise disjoint, for a fixed m and that

$$\sum_{i=1}^{m} P(A_{(i,m)}) = P(\bigcup_{i=1}^{m} A_{(i,m)}) = 1$$

I don't really understand why all the $P(A_{(i,m)})$'s are equal

For example why is it that $$P(A_{1,2}) := P(Y_1 \ge Y_2) = P(Y_2 \ge Y_1) := P(A_{2,2})$$ Why 1/2 each? Why not 1/4, 3/4? Or even 1, 0? I'm guessing this has something to do w/ independence or continuity.

Assuming the random variables are absolutely continuous (What if the pdfs don't exist? :O),

$$P(B_m) = \int_{\mathbb R} \int_{-\infty}^{y_n} \cdots \int_{-\infty}^{y_n} \int_{-\infty}^{y_n} f_{Y_1, ..., Y_n} (y_1, ..., y_n) dy_1dy_2...dy_{n-1}dy_n$$

By independence, we have

$$= \int_{\mathbb R} \int_{-\infty}^{y_n} \cdots \int_{-\infty}^{y_n} \int_{-\infty}^{y_n} f_{Y_1}(y_1) \cdots f_{Y_n}(y_n) dy_1dy_2...dy_{n-1}dy_n$$

$$= \int_{\mathbb R} [\int_{-\infty}^{y_n} \cdots \int_{-\infty}^{y_n} \int_{-\infty}^{y_n} f_{Y_1}(y_1) \cdots f_{Y_{n-1}}(y_{n-1}) dy_1dy_2...dy_{n-1}] f_{Y_n}(y_n) dy_n$$

$$= \int_{\mathbb R} [F_{Y_1}(y_n) ... F_{Y_{n-1}}(y_n)] f_{Y_n}(y_n) dy_n$$

By identical distribution,

$$= \int_{\mathbb R} [F_{Y_n}(y_n) ... F_{Y_{n}}(y_n)] f_{Y_n}(y_n) dy_n$$

$$= \int_{\mathbb R} [F_{Y_n}(y_n)]^{n-1} f_{Y_n}(y_n) dy_n$$

$$= [F_{Y_n}(y_n)]^{n-1} F_{Y_n}(y_n)|_{-\infty}^{\infty} - \int_{\mathbb R} [F_{Y_n}(y_n)]^{n-1} (n-1) f_{Y_n}(y_n) dy_n$$

$$= [(1)(1) - (0)(0)] - \int_{\mathbb R} [F_{Y_n}(y_n)]^{n-1} (n-1) f_{Y_n}(y_n) dy_n$$

$$= 1/n$$

• It is true and is a direct consequence of the fact that $Y_1,\dots,Y_m$ are iid and continuously distributed. E.g. $P(Y_1>Y_2)=P(Y_2>Y_1)$ follows by symmetry and we have $P(Y_1>Y_2)+P(Y_2>Y_1)=1-P(Y_1=Y_2)=1-0=1$. Consequently $P(Y_1>Y_2)=P(Y_2>Y_1)=\frac12$ – drhab Nov 21 '15 at 15:00
• Wonder why in this context e.g $Y_3$ would have more or less chance to be the largest of $Y_1,\dots,Y_m$ than e.g. $Y_5$. No reason can be found. If the distribution is not continuous then difficulties arise because it can happen that more than one rv can be the maximum. – drhab Nov 21 '15 at 15:05
• Symmetry gives $P(f(Y_1,Y_2)\in A)=P(f(Y_2,Y_1)\in A)$. You can simply switch. This in any expression. This because $(Y_1,Y_2)$ and $(Y_2,Y_1)$ have the same distribution. More generally $P\left(f\left(\langle Y_{1},\dots,Y_{m}\rangle\right)\in A\right)=P\left(f\left(\langle Y_{\sigma\left(1\right)},\dots,Y_{\sigma\left(m\right)}\rangle\right)\in A\right)$ where $\sigma$ is a permutation. – drhab Nov 21 '15 at 15:43
• @drhab Thanks. How do you know that? May you pls provide a link or reference of some kind? Btw is the small f a PDF? How do you know the pdf s exist? – BCLC Nov 23 '15 at 2:51
• In my comment $f$ is not a PDF. You can take any measurable function for it. E.g. $f(x,y)=x-y$, Then for $A=(0,\infty)$ you get $P(Y_1>Y_2)=P(Y_2>Y_1)$. If $Y_1,Y_2$ are iid then $(Y_1,Y_2)$ and $(Y_2,Y_1)$ have the same distribution: $F_{Y_1,Y_2}(u,v)=F(u)F(v)=F_{Y_2,Y_1}(u,v)$ where $F=F_{Y_1}=F_{Y_2}$. That is all you need. – drhab Nov 23 '15 at 11:09

If $Y_1,\dots,Y_n$ are iid and $\sigma:\{1,\dots,n\}\to\{1,\dots,n\}$ is a permutation then: $$F_{Y_{\sigma(1)},\dots Y_{\sigma(n)}}(y_1,\dots,y_n)=F(y_1)\times\cdots\times F(y_n)=F_{Y_1,\dots,Y_n}(y_1,\dots,y_n)$$ where $F$ is the common CDF of the $Y_i$.

So the random vectors $\langle Y_{\sigma(1)},\dots,Y_{\sigma(n)}\rangle$ and $\langle Y_1,\dots,Y_n\rangle$ have equal distribution.

Consequently the events $\{Y_n=\max(Y_1,\dots,Y_n)\}$ and $\{Y_{\sigma(n)}=\max(Y_1,\dots,Y_n)\}$ have equal probability.