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!

 

[Ray Vickson]

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.

 

[Ray Vickson]

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 $P(X_1 < 2X_2)$.

 

[rayge]

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

 

[Ray Vickson]

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

Please: no guesses. Go back to your probability textbook and read the appropriate sections, or consult your course notes. This is all really basic material about bivariate distributions, etc.