# Find probability by conditioning on a random variable.

If $X_1,X_2,\ldots,X_n$ are independent and identically distributed continuous random variables, show that $P(X_1<X_2,X_1<X_3,\ldots,X_1<X_n)=\frac{1}{n}$.

So far, this is what I have: $$P(X_1<X_2,X_1<X_3,\ldots,X_1<X_n)=\int_{-\infty}^{\infty}P(X_1<X_2,X_1<X_3,\ldots,X_1<X_n|X_1=x)\cdot f_{X_1}(x)dx$$ $$=\int_{-\infty}^{\infty}P(X_1<x,X_1<x,\ldots,X_1<x)\cdot f_{X_1}(x)dx$$ $$=\int_{-\infty}^{\infty}\prod_{i=1}^{n}P(X_i>x)\cdot f_{X_1}(x)dx$$

This is where I'm stuck. I know that since they're all identically distributed, there is probably some way to put everything together that is inside the integral, but I'm having a hard time figuring out how $P(X_i>x)$ relates to the distribution of $X_1$.

• Consider that $Pr(X_i>x) = 1 - Pr(X_i \le x) =1- F(x)$, where $F(x)$ is the cdf of $X_i$ and $F'(x)=f_{X_i}(x)$. – user477602 Oct 1 '17 at 16:08
• @Opt Then I get $\int_{-\infty}^{\infty} \prod_{i=1}^{n}[1-F(x_i)]f_{X_1}(x) dx$ and I'm not sure how to deal with the product. – mmm Oct 1 '17 at 18:58

Let $F$ be the CDF of $X_1$. Then $$\mathsf{P}\left(\bigcap_{i> 1}\{X_1<X_i\}\mid X_1=x\right)=\prod_{i>1}\mathsf{P}(X_i>x)=(1-F(x))^{n-1}$$ and $$\mathsf{P}\left(\bigcap_{i> 1}\{X_1<X_i\}\right)=\int (1-F(x))^{n-1}dF(x)=-\frac{1}{n}(1-F(x))^n\mid_{-\infty}^{+\infty}=\frac{1}{n}.$$