I have $n+1$ i.i.d. RV $X_1,\dots,X_n,X$. I compute the order statistics of $X_1,\dots,X_n$:
$$X_{(1)},\dots,X_{(n)}$$
I'm reading a paper which says that $P(X \leq X_{(k)})=\frac{k}{n+1}$ (see Appendix 1, theorem A.1). How can I prove it? For the sake of simplicity, it's ok to assume that there are no ties, or that the $X_i$ are continuous RV (which implies that there are no ties with probability 1).
As a start, I tried to compute $P(X \leq X_{(1)})$:
$$P(X \leq X_{(1)})=P(X \leq X_1,\dots,X\leq X_n)=P(X \leq X_1)\dots P(X\leq X_n)$$
because of independence. Now, since $X_1,\dots,X_n,X$ are i.d., I just need to compute $P(X \leq X_1)$: denoting with $f(x), F(x)$ respectively the pdf and the CDF of $X$, this is
$$\int_{-\infty}^{\infty}\int_{-\infty}^{x_1}p(x,x_1)dxdx_1=\int_{-\infty}^{\infty}\int_{-\infty}^{x_1}f(x)f(x_1)dxdx_1=\int_{-\infty}^{\infty}f(x_1)\left(\int_{-\infty}^{x_1}f(x)dx\right)dx_1=\int_{-\infty}^{\infty}f(x_1)F(x_1)dx_1=[F(x)F(x)]_{-\infty}^{\infty}-\int_{-\infty}^{\infty}f(x_1)F(x_1)dx_1=1-\int_{-\infty}^{\infty}f(x_1)F(x_1)dx_1\implies\int_{-\infty}^{\infty}\int_{-\infty}^{x_1}p(x,x_1)dxdx_1=\frac12$$
Thus, I would have that
$$P(X \leq X_{(1)})=P(X \leq X_1,\dots,X\leq X_n)=P(X \leq X_1)\dots P(X\leq X_n)=\frac{1}{2^n}$$
which is clearly not what the paper claims. Where did I go wrong, and what is the correct proof?
Requested from comments:
As Exodd commented, the events $X\le X_i$ are not independent, so you cannot take the product of their probabilities.
Assuming ties have probability $0$, there is a simple combinatorial argument to answer your question:
there are $(n+1)!$ equally likely possible orderings of $X_1,\ldots,X_n,X$;
so $X$ is equally likely to be in any of the $n+1$ positions overall;
so $X$ is one of the smallest $k$ values with probability $\frac{k}{n+1}$;
so $P(X \leq X_{(k)})=\frac{k}{n+1}$.