Correctly expressing conditional cdf

Question

The question is:

Given the cdf $F_X(x)$, find $F_X(x\mid X\gt t)$ and $f_X(x\mid X\gt t)$ for some $t\in\mathbb{R}$.

Since $f_X(x\mid X\gt t) = \frac{\partial F_X(x\mid X\gt t)}{\partial x}$, we just need to find $F_X(x\mid X\gt t) = \frac{\mathbb{P}(\{X\leq x\}\cap\{X\gt t\})}{\mathbb{P}(X\gt t)} = \frac{\mathbb{P}(\{X\leq x\}\cap\{X\gt t\})}{1 - F_X(t)}$.

The term $\mathbb{P}(\{X\leq x\}\cap\{X\gt t\})$ can be calculated in three ways (that occur to me):

• $\mathbb{P}(\{X\leq x\}\cap\{X\gt t\}) = \mathbb{P}(t\lt X \leq x) = $ $\begin{cases} 0 & x\lt t \\ F_X(x)-F_X(t) & x\geq t \end{cases}$
• $\mathbb{P}(\{X\leq x\}\cap\{X\gt t\}) = \mathbb{P}(t\lt X \leq x) = \mathbb{P}(X\gt t)\cdot \mathbb{P}(X \leq x\mid X\gt t)$
• $\mathbb{P}(\{X\leq x\}\cap\{X\gt t\}) = \mathbb{P}(t\lt X \leq x) = \mathbb{P}(X\leq x)\cdot \mathbb{P}(X\gt t\mid X \leq x)$

The first method is the most straightforward and using that we get $$F_X(x\mid X\gt t) = \begin{cases} 0 & x\lt t \\ \frac{F_X(x)-F_X(t)}{1-F_X(t)} & x\geq t \end{cases}$$ I understood this, but I initially tried the problem using the last two methods and got stuck and I haven't able to track my mistake.

I considered the events $\{X\gt t\}$ and $\{X\leq x\}$ in $\mathbb{P}(X \leq x\mid X\gt t)$ and $\mathbb{P}(X\gt t\mid X \leq x)$ to be independent if and only if $x\geq t$ and dependent otherwise. That is,

• $\mathbb{P}(X \leq x\mid X\gt t) = $ $\begin{cases} 0 & x\lt t \\ \frac{F_X(x)\cdot(1-F_X(t))}{1-F_X(t)} = F_X(x) & x\geq t \end{cases}$
• $\mathbb{P}(X\gt t\mid X \leq x) = $ $\begin{cases} 0 & x\lt t \\ \frac{F_X(x)\cdot(1-F_X(t))}{F_X(x)} = 1-F_X(t) & x\geq t \end{cases}$

Using this we get (in order),

• $F_X(x\mid X\gt t) = \frac{\mathbb{P}(X\gt t)\cdot \mathbb{P}(X \leq x\mid X\gt t)}{1-F_X(t)} = \begin{cases} 0 & x\lt t \\ \frac{F_X(x)\cdot(1-F_X(t))}{1-F_X(t)} = F_X(x) & x\geq t \end{cases}$
• $F_X(x\mid X\gt t) = \frac{\mathbb{P}(X\leq x)\cdot \mathbb{P}(X \gt t\mid X\leq x)}{1-F_X(t)} = \begin{cases} 0 & x\lt t \\ \frac{F_X(x)\cdot(1-F_X(t))}{1-F_X(t)} = F_X(x) & x\geq t \end{cases}$

They turn out to be the same, but wrong! My guess is that my assumption of independence is incorrect, but at the same time, I'm not able to justify why or suggest an alternative.

Any help will be appreciated, thanks!

Answer 1

First of all, the events $\{X>t\}$ and $\{X\leq x\}$ are independent if only if at least one of its probabilities is $0$ or $1$. Indeed, $$ \mathbb P(X>t, X\leq x) = \mathbb P(t < X\leq x)=\begin{cases}F_X(x)-F_X(t), & x>t \\ 0, & x \leq t\end{cases} \tag{1} $$ can be equal to $$ \mathbb P(X>t)\mathbb P(X\leq x)= (1-F_X(t))F_X(x) = F_X(x) - F_X(t)F_X(x) \tag{2} $$

for $x>t$: iff $F_X(t)=F_X(t)F_X(x)$ $\iff$ $F_X(x)=1$ or $F_X(t)=0$,

for $x\leq t$: iff $F_X(x)=0$ or $F_X(t)=1$.

So your calculations for second and third cases are wrong. Moreover: what is the need to express conditional probability through itself? Look carefully to the second step. You need to find initially $\mathbb P(X\leq x\mid X>t)$. You rewrite it as $$ \color{red}{\mathbb P(X\leq x\mid X>t)} = \frac{\mathbb P(X\leq x, X>t)}{\mathbb P(X>t)}= \frac{\mathbb P(X > t)\cdot \color{red}{\mathbb P(X \leq x\mid X > t)}}{\mathbb P(X>t)}=\color{red}{\mathbb P(X \leq x\mid X > t)} $$ What's the point of this?