Distribution of R.V. X given X < Y (or > Y) and Y > constant k

Question

I am looking for the value of two expressions (expression \ref{star} and expression \ref{circ}), both of which are quite basic, and both of which I am ashamed to admit that I am getting wrong.

Suppose there are two continuously distributed independent random variables, $X$ and $Y$, whose respective distributions $F$ and $G$ each have support on $[0,1]$. Moreover, suppose both admit densities.

First, I'd like, \begin{equation}\tag{$\star$}\label{star}\Pr(X < x | X > Y, Y > k)\end{equation} where $k \in [0,1]$. To derive this expression I do

$$\begin{split}\Pr(X < x | X > Y, Y > k) &= \frac{\Pr(X < x \cap X > Y \cap Y > k)}{\Pr(X > Y \cap Y > k)}\end{split}$$ The numerator is $$\begin{split}\Pr(X < x \cap X > Y \cap Y > k) &= \frac{1}{1-G(k)}\int_{k}^{1}\Pr(X < x \cap X > Y|Y=y)g(y)dy\\ &= \frac{1}{1-G(k)}\int_{k}^{1}\Pr(X < x \cap X > y)g(y)dy\end{split}$$ since $X$ and $Y$ are independent. Next, $$\Pr(X < x \cap X > y) = \begin{cases}F(x) - F(y), & y \leq x\\ 0, & y > x\end{cases}$$ Hence, $$\begin{split}\Pr(X < x \cap X > Y \cap Y > k) &= \frac{1}{1-G(k)}\int_{k}^{x}\left[F(x) - F(y)\right]g(y)dy\\ &= \frac{1}{1-G(k)}\left\{F(x)\left[G(x) - G(k)\right] - \int_{k}^{x}F(y)g(y)dy\right\}\end{split}$$ with support on $[k,1]$. Evaluated at $x=k$, this gives $0$, as required. However, evaluated at $x = 1$ this gives

$$1 - \frac{\int_{k}^{1}F(y)g(y)dy}{1-G(k)} < 1$$

This is Incorrect. The denominator is

$$\begin{split}\Pr(Y < X \cap Y > k) = \frac{1}{1-G(k)}\int_{k}^{1}\left(1-F(y)\right)g(y)dy\end{split}$$ which I believe is correct, though this of course may be wrong as well.

Second, I'd like, $$\label{circ}\tag{$\circ$}\Pr(X < x | X < Y, Y > k)$$ where $k \in [0,1]$. As above,

$$\begin{split}\Pr(X < x | X < Y, Y > k) &= \frac{\Pr(X < x \cap X < Y \cap Y > k)}{\Pr(X < Y \cap Y > k)}\end{split}$$ The numerator is $$\begin{split}\Pr(X < x \cap X < Y \cap Y > k) &= \frac{1}{1-G(k)}\int_{k}^{1}\Pr(X < x \cap X < Y|Y=y)g(y)dy\\ &= \frac{1}{1-G(k)}\int_{k}^{1}\Pr(X < x \cap X < y)g(y)dy\end{split}$$ since $X$ and $Y$ are independent. Next, $$\Pr(X < x \cap X < y) = \begin{cases}F(y), & y \leq x\\ F(x), & y > x\end{cases}$$ Hence, $$\begin{split}\Pr(X < x \cap X > Y \cap Y > k) &= \frac{1}{1-G(k)}\left[\int_{k}^{x}F(y)g(y)dy + \int_{x}^{1}F(x)g(y)dy\right]\\ &= \frac{1}{1-G(k)}\left\{\int_{k}^{x}F(y)g(y)dy + F(x)\left[1-G(x)\right]\right\}\end{split}$$ on $[k,1]$ and $F(x)$ on $[0,k]$. Evaluated at $x=0$, this gives $0$, as required, evaluated at $x=k$, this gives $F(k)$, also good. However, evaluated at $x = 1$ this gives

$$1 + \frac{\int_{k}^{1}F(y)g(y)dy}{1-G(k)} > 1$$

This is also Incorrect. The denominator is

$$\begin{split}\Pr(Y > X \cap Y > k) = F(k) + \frac{1}{1-G(k)}\int_{k}^{1}F(y)g(y)dy\end{split}$$ which I believe is correct, though this of course may be wrong as well.

Answer 1

$\begin{align}\mathsf P(X<x\mid k<Y<X)&=\dfrac{ \mathsf P(k<Y<X<x)}{\mathsf P(k<Y<X)}\\[2ex] &= \dfrac{\int_k^x \mathsf P(t<X<x\mid Y=t)~G'(t)~\mathsf d t}{\int_k^1\mathsf P(t<X\mid Y=t)~G'(t)~\mathsf d t}\\[2ex] &= \dfrac{\int_k^x \mathsf P(t<X<x)~G'(t)~\mathsf d t}{\int_k^1\mathsf P(t<X)~G'(t)~\mathsf d t}\\[2ex]&=\dfrac{\int_k^x \left(F(x)-F(t)\right)~G'(t)~\mathsf d t}{\int_k^1 \left(F(1)-F(t)\right)~G'(t)~\mathsf d t}\\[2ex]&=\dfrac{F(x)~(G(x)-G(k))-\int_k^x F(t)~G'(t)~\mathsf d t}{F(1)~(G(1)-G(k))-\int_k^1 F(t)~G'(t)~\mathsf d t}\\[2ex]&=\dfrac{F(x)~(G(x)-G(k))-\int_k^x F(t)~G'(t)~\mathsf d t}{(1-G(k))-\int_k^1 F(t)~G'(t)~\mathsf d t}\end{align}$

Answer 2

One mistake I see is you are writing $P(A\cap B)=\frac {P(A|B)} {P(B)}$ whereas the correct formula is $P(A\cap B)=P(A|B) P(B)$. Thus $1-G(k)$ should not be in the denominator.