Joint distribution of random variable with an order statistic

Question

Let $X_1,...,X_n$ be i.i.d. with the distribution of the $X_i$ being nice and continuous. I'm interested in the expression of the CDF $F_{X_{(1)},X_j}(u,v)$. To be clear $X_{(1)} = min(X_i)$ and $X_j$ just one of the n i.i.d. random variables.

I'm wondering if the following derivation is correct?

$$F_{X_{(1)},X_j}(u,v) = P(X_{(1)}\leq u, X_j\leq v)$$

Partition the probability by whether or not $X_j$ is the minimum $$=P(X_{(1)}\leq u,X_j\leq v,X_{(1)}= X_j)+P(X_{(1)}\leq u,X_j\leq v,X_{(1)}\neq X_j)$$

The first probability reduces down as follows

$$P(X_{(1)}\leq u,X_j\leq v,X_{(1)}= X_j)$$ $$=P(X_{(1)}\leq min(u,v), X_{(1)}= X_j)$$

The above is the probability that $X_j$ is minimum of the n i.i.d. random variables and it is less than both $u$ and $v$. Since the distribution of the $X_i$ is continuous we can compute this directly as $$=\int_{-\infty}^{min(u,v)}{f_X(x)[1-F_X(x)]^{n-1}dx}$$ $$=\frac{1-[1-F_X(min(u,v))]^{n}}{n}$$

The last equality comes from integrating by substitution.

Returning to the other probability we have $$P(X_{(1)}\leq u,X_j\leq v,X_{(1)}\neq X_j)$$ $$=P(X_{(1)}\leq min(u,v),X_j\leq v,X_{(1)}\neq X_j)$$

Since the $X_i$ are i.i.d., the above is equivalent to the probability that $$P(X_j\leq v, X'_{(1)}\leq min(u,v),X'_{(1)}\leq X_j )$$ where $X'_{(1)}$ is the minimum of the other $n-1$ random variables, so that $X_j$ and $X'_{(1)}$ are independent. The probability above can be written as $$\int_{-\infty}^{min(u,v)}{[F_X(v)-F_X(x)]f_{X'_{(1)}}(x)dx}$$

Using the fact that $f_{X'_{(1)}}(x)=(n-1)f_X(x)[1-F_X(x)]^{n-2}$ we have the the above probability can be computed as

$$\int_{-\infty}^{min(u,v)}{[F_X(v)-F_X(x)](n-1)f_X(x)[1-F_X(x)]^{n-2}dx}$$

To compute the above integral, let $s=min(u,v), b=F_X(v),$ and use the substitution $y=1-F_X(x)$ so that the above integral becomes

$$\int_{1}^{s}{-(n-1)(b-1+y)(y)^{n-2}dy}$$ $$=-(n-1)\int_{1}^{s}{[(b-1)y^{n-2}+y^{n-1}]dy}$$ $$=-(n-1)\int_{1}^{s}{[(b-1)y^{n-2}+y^{n-1}]dy}$$ $$=(n-1)\int_{1}^{s}{[(1-b)y^{n-2}-y^{n-1}]dy}$$ $$=(1-b)y^{n-1}-y^{n}\frac{(n-1)}{n}\Big|_1^s$$

$$=[1-F_X(v)]([1-F_X(s)]^{n-1}-1)+\frac{(n-1)}{n}(1-[1-F_X(s)]^{n})$$

So, both parts together give us $$=\frac{1-[1-F_X(s)]^{n}}{n}+[1-F_X(v)]([1-F_X(s)]^{n-1}-1)+\frac{(n-1)(1-[1-F_X(s)]^{n})}{n}$$ $$=[1-F_X(v)]([1-F_X(s)]^{n-1}-1)+(1-[1-F_X(s)]^{n})$$

and finally, with $s=min(u,v)$:

$$F_{X_{(1)},X_j}(u,v) = [1-F_X(v)]([1-F_X(s))]^{n-1}-1)+(1-[1-F_X(s)]^{n})$$

Answer 1

It is obvious that for every $j\in\{1,\dots,n\}$:$$P\left(X_{\left(1\right)}\leq u,X_{j}\leq v\right)=P\left(X_{\left(1\right)}\leq u,X_{1}\leq v\right)$$

If $v\leq u$ then $\left\{ X_{\left(1\right)}\leq u,X_{1}\leq v\right\} =\left\{ X_{1}\leq v\right\} $ so that in that case: $$P\left(X_{\left(1\right)}\leq u,X_{1}\leq v\right)=P\left(X_{1}\leq v\right)=F_{X}\left(v\right)$$

If $v>u$ then we can go for:

$$\begin{aligned}P\left(X_{\left(1\right)}\leq u,X_{1}\leq v\right) & =P\left(X_{\left(1\right)}\leq u,X_{1}\leq u\right)+P\left(X_{\left(1\right)}\leq u,u<X_{1}\leq v\right)\\ & =P\left(X_{1}\leq u\right)+P\left(\min\left(X_{2},\dots,X_{n}\right)\leq u,u<X_{1}\leq v\right)\\ & =P\left(X_{1}\leq u\right)+P\left(\min\left(X_{2},\dots,X_{n}\right)\leq u\right)P\left(u<X_{1}\leq v\right)\\ & =F_{X}\left(u\right)+\left(1-\left(1-F_{X}\left(u\right)\right)^{n-1}\right)\left(F_{X}\left(v\right)-F_{X}\left(u\right)\right) \end{aligned} $$

Answer 2

From @drhab's answer we see that the joint distribution of $(X_{(1)},X_j)$ is given by $$ G(u,v) = F(v)\mathsf 1_{\{v\leqslant u\}} + \left(1-(1-F(u))^{n-1}\right)(F(v)-F(u))\mathsf 1_{\{v>u\}}. $$ For a concrete example, let $X_n\stackrel{\mathrm{i.i.d.}}{\sim}\mathrm{Expo}(\lambda)$, that is, $F(t) = \left(1 - e^{-\lambda t}\right)\mathsf 1_{(0,\infty)}(t)$. Then the joint distribution of $(X_{(1)},X_j)$ is given by \begin{align} G(u,v) =& F(v)\mathsf 1_{\{v\leqslant u\}} + \left(1-(1-F(u))^{n-1}\right)(F(v)-F(u))\mathsf 1_{\{v>u\}}\\ &=\begin{cases} 1-e^{-\lambda v},& v\leqslant u\\ \left(1-e^{-(n-1)\lambda u}\right)\left(e^{-\lambda u}-e^{-\lambda v}\right),& v>0. \end{cases} \end{align}