Probability inequality of multiple continuous random variables Let $X_0, \ldots , X_{n-1}$ be i.i.d positive random variables with a continuous probability function $f_X$. I'm trying to prove that
$$
\mathbb{P}[X_0 \ge max(X_1, \ldots , X_{n-1})] = \int_0^{\infty} (F_{X_0})^{n-1}(a)f_{X_0}(a) \,da
$$
Where $F$ is the CDF and $f$ is the PDF
I guess that $\mathbb{P}[X_0 \ge max(X_1, \ldots , X_{n-1})]= (n-1) \mathbb{P}[X_0 \ge X_1]$ because they're i.d.d, but how do I translate it to the CDF?
 A: Your guess is wong. 
Realize that $\mathbb P(X_0\geq X_1)=\frac12$ on base of:


*

*$P(X_0\geq X_1)+P(X_1\geq X_0)-P(X_0=X_1)$

*$P(X_0\geq X_1)=P(X_1\geq X_0)$ on base of symmetry.

*$P(X_0=X_1)=0$ on base of continuity.


Likewise it can be verified that: $$P\left(X_0\geq\max(X_1,\dots,X_{n-1})\right)=\frac1n$$

edit:
Also $$P\left(X_0\geq\max(X_1,\dots,X_{n-1})\right)=\int P\left(X_0\geq\max(X_1,\dots,X_{n-1})\mid X_0=a\right)f_X(a)da=$$$$\int P(\max(X_1,\dots,X_{n-1})\leq a)f_X(a)da=\int F_X^{n-1}(a)f_X(a)da$$
where the second equality rests on independence.
A: Disclaimer: this is just a remark on drhab's answer.
Note that
\begin{align}
(F_{X_0})^{n-1}(a)f_{X_0}(a) = \frac{d}{da} \frac{1}{n} (F_{X_0})^{n}(a).
\end{align}
The fundamental theorem of calculus implies that the integral is equal to
\begin{align}
\lim_{a \to \infty } \frac{1}{n} (F_{X_0})^{n}(a) - \frac{1}{n} (F_{X_0})^{n}(0) = \frac{1}{n},
\end{align}
where we used that $X_0$ is a positive random variable.
A: Since the variables are i.i.d. it is
\begin{align*}
\mathbb{P}(X_0\geq \max(X_1,\ldots,X_{n-1})|X_0=x) &= \mathbb{P}(x\geq \max(X_1,\ldots,X_{n-1})|X_0=x) \\
& = \mathbb{P}(x\geq \max(X_1,\ldots,X_{n-1})) \\
& = \mathbb{P}(X_1\leq x,\ldots,X_{n-1}\leq x) \\
& = \mathbb{P}(X_1\leq x)^{n-1} \\
& = F_{X_0}(x)^{n-1}.
\end{align*}
Hence:
\begin{align*}
\mathbb{P}(X_0\geq \max(X_1,\ldots,X_{n-1})) &=\int_0^{+\infty}\mathbb{P}(X_0\geq \max(X_1,\ldots,X_{n-1})|X_0=x)f_{X_0}(x)dx \\
&=\int_0^{+\infty} F_{X_0}(x)^{n-1}f_{X_0}(x)dx
\end{align*}
Now you can use the fact that $F_{X_0}(x)^{n-1}f_{X_0}=\frac{1}{n}\frac{d}{dx}F_{X_0}(x)^{n}$ to show that $\mathbb{P}(X_0\geq \max(X_1,\ldots,X_{n-1}))=\frac{1}{n}$.
Another way to reach this last result is the following. Let $X_{(n)}=\max(X_0,X_1,\ldots,X_{n-1})$. The event $\{X_0\geq \max(X_1,\ldots,X_{n-1})\}$ is equivalent to the event $\{X_0=X_{(n)}\}$. Since the variables are i.i.d., $\mathbb{P}(X_0=X_{(n)})=\mathbb{P}(X_i=X_{(n)})=\mathbb{P}(X_i \textrm{ is the largest among }X_0,\ldots,X_{n-1})$ for all $i=0,\ldots,n-1$. Additionally, since ties have probability zero (as the distributions are continuous) it is
$$1=\sum_{i=0}^{n-1} \mathbb{P}(X_i=X_{(n)}) = n\mathbb{P}(X_0=X_{(n)})=n\mathbb{P}(X_0\geq \max(X_1,\ldots,X_{n-1}))$$ 
