Finding Conditional PDF and Conditional CDF

Question:

Let $X$ be a continuous random variable with CDF $F$ and PDF $f$

(a)Find the conditional CDF of $X$ given $X>a$ (where $a$ is a constant with $P(X>a) \neq 0$). That is, find $P(X \leq x | X \gt a)$, in terms of $F$.

(b) Find the conditional PDF of $X$ given $X \gt a$.

(c) Check that the conditional PDF from (b) is a valid PDF, showing directly that it is nonnegative and integrates to 1.

I have this much so far:

So the answer to (a) is $\frac{F(x)-F(a)}{1-F(a)}$. I don't understand how to proceed further. Can someone tell me if my first answer is correct and help me proceed?

• Your first answer looks fine. Now that you have a CDF, how do you find a PDF? (Don't let the word "conditional" scare you; you just have another probability distribution which happens to be on the interval $(a, \infty)$.) Once you find the PDF, check that it is nonnegative and integrates to $1$. – angryavian Oct 26 '17 at 6:45
• Do we find the derivative of the CDF? – user490867 Oct 26 '17 at 6:47
• Sorry but is "your first answer" really yours, or did you copy it somewhere? I am asking because its meaning seems to elude you completely... – Did Oct 26 '17 at 8:55

It is: $$G(x):=\frac{F(x)-F(a)}{1-F(a)}\times\chi_{(a,\infty)}(x)$$ where $\chi_{(a,\infty)}$ denotes the indicator function of set $(a,\infty)$.

In lots of cases your comment concerning finding PDF can be answered with: "yes, just take the derivative."

That would lead to: $$g(x):=\frac{f(x)\chi_{(a,\infty)}(x)}{1-F(a)}\tag1$$ where $\chi_{(a,\infty)}$ denotes the indicator function of set $(a,\infty)$.

It is immediate that $g(x)$ takes only non-negative values.

But this of course only works if $F$ has a indeed derivative.

There is a more elementary way to verify that $(1)$ works as PDF here.

You only have to verify that $$\int_{-\infty}^xg(u)du=G(x)\tag2$$

From $(2)$ it also follows directly that $$\int_{-\infty}^{\infty}g(u)du=\lim_{n\to\infty}G(x)=1$$