# Conditital probability of a random variable.

• Mar 22nd 2011, 11:18 AM
Sheld
Conditital probability of a random variable.
I given that the pdf of a continuous random variable X is

$\displaystyle f(x) = e^{-x}$ for $\displaystyle x > 0$ and $\displaystyle 0$ for $\displaystyle x \leq 0$

I have to find $\displaystyle P(X \leq 2 | X > 1)$
Am I correct in thinking that

$\displaystyle P((X \leq 2) \cap (X > 1)) = P(1 < X \leq 2)$ ?
• Mar 22nd 2011, 11:39 AM
dwsmith
Quote:

Originally Posted by Sheld
I given that the pdf of a continuous random variable X is

$\displaystyle f(x) = e^{-x}$ for $\displaystyle x > 0$ and $\displaystyle 0$ for $\displaystyle x \leq 0$

I have to find $\displaystyle P(X \leq 2 | X > 1)$
Am I correct in thinking that

$\displaystyle P((X \leq 2) \cap (X > 1)) = P(1 < X \leq 2)$ ?

Yes.
• Mar 22nd 2011, 01:02 PM
Sheld
Thanks for answer. I have another quick question, when they say find the distribution of a random variable X, they mean the cumulative distribution function, right?
• Mar 22nd 2011, 01:47 PM
dwsmith
Quote:

Originally Posted by Sheld
Thanks for answer. I have another quick question, when they say find the distribution of a random variable X, they mean the cumulative distribution function, right?

I think so.
• Mar 22nd 2011, 04:22 PM
matheagle
Quote:

Originally Posted by Sheld
Thanks for answer. I have another quick question, when they say find the distribution of a random variable X, they mean the cumulative distribution function, right?

It can mean either the cumulative or the density.
In this case I would say, it's an exponential with mean, beta=1.