# Evaluating Conditional Probability of Several Random Variables

rayge

## Homework Statement

Let $X_1, X_2, X_3$ be iid with common pdf $f(x)=exp(-x)$, $0<x<infinity$, $0$ elsewhere.

Evaluate $P(X_1<X_2 | X_1<2X_2)$

## Homework Equations

$f(X|Y) = f(x,y)/f(y)$

## The Attempt at a Solution

Since $P(X_1<X_2)$ is a subset of $P(X_1<2X_2)$, the intersection (edited, at first said union) should be $P(X_1<X_2)$, so the conditional should be $P(X_1<X_2)/P(X_1<2X_2)$ (edited). I evaluate this and get $(exp(-x_2) - 1 )/ (exp(-2x_2 ) - 1)$, so I'm going wrong somewhere. Any suggestions welcome!

Last edited:

Homework Helper
Dearly Missed

## Homework Statement

Let $X_1, X_2, X_3$ be iid with common pdf $f(x)=exp(-x)$, $0<x<infinity$, $0$ elsewhere.

Evaluate $P(X_1<X_2 | X_1<2X_2)$

## Homework Equations

$f(X|Y) = f(x,y)/f(y)$

## The Attempt at a Solution

Since $P(X_1<X_2)$ is a subset of $P(X_1<2X_2)$, the union should be $P(X_1<X_2)$, so the conditional should be $P(X_1<2X_2)/P(X_1<2X_2)$. I evaluate this and get $(exp(-x_2) - 1 )/ (exp(-2x_2 ) - 1)$, so I'm going wrong somewhere. Any suggestions welcome!

What value has ##x_2?## There is no ##x_2## mentioned in the question---only the random variable ##X_2##. Also, you should look at an intersection, not a union: the intersection is ##\{X_1 < X_2\}## while the union is ##\{X_1 < 2X_2\}##. Furthermore, you say one thing and compute something else.

Also, what role does ##X_3## have in the question?

rayge
Edited, you are right. I had evaluated $P(X_1<X_2)/P(X_1<2X_2)$ to get the answer.

$X_3$ is the second part of the problem, not used in this part. Sorry for the confusion.

Homework Helper
Dearly Missed
Edited, you are right. I had evaluated $P(X_1<X_2)/P(X_1<2X_2)$ to get the answer.

$X_3$ is the second part of the problem, not used in this part. Sorry for the confusion.

No, that is not what you calculated. You calculated
$$\frac{P(X_1 < x_2)}{P(X_1 < 2x_2)}$$
for some un-specified value of the number ##x_2##.

How do you calculate ##P(X_1 < X_2)##? Here, I wrote and I mean ##X_2##---the random variable--not a number ##x_2##. Same question for ##P(X_1 < 2X_2)##.

Last edited:
rayge
My guess is that it's the probability that $X_1$ is less than the mean of $X_2$, not some $x_2$.

My guess is that it's the probability that $X_1$ is less than the mean of $X_2$, not some $x_2$.