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&lt;X_2)/P(X_1&lt;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&lt;X_2)/P(X_1&lt;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 &lt; x_2)}{P(X_1 &lt; 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.