Evaluating Conditional Probability of Several Random Variables

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  • #1
rayge
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Homework Statement


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

Evaluate [itex]P(X_1<X_2 | X_1<2X_2)[/itex]


Homework Equations


[itex]f(X|Y) = f(x,y)/f(y)[/itex]

The Attempt at a Solution


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

  • #2
Ray Vickson
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Homework Statement


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

Evaluate [itex]P(X_1<X_2 | X_1<2X_2)[/itex]


Homework Equations


[itex]f(X|Y) = f(x,y)/f(y)[/itex]

The Attempt at a Solution


Since [itex]P(X_1<X_2)[/itex] is a subset of [itex]P(X_1<2X_2)[/itex], the union should be [itex]P(X_1<X_2)[/itex], so the conditional should be [itex]P(X_1<2X_2)/P(X_1<2X_2)[/itex]. I evaluate this and get [itex](exp(-x_2) - 1 )/ (exp(-2x_2 ) - 1)[/itex], 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?
 
  • #3
rayge
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Edited, you are right. I had evaluated [itex]P(X_1<X_2)/P(X_1<2X_2)[/itex] to get the answer.

[itex]X_3[/itex] is the second part of the problem, not used in this part. Sorry for the confusion.
 
  • #4
Ray Vickson
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Edited, you are right. I had evaluated [itex]P(X_1<X_2)/P(X_1<2X_2)[/itex] to get the answer.

[itex]X_3[/itex] 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
[tex] \frac{P(X_1 < x_2)}{P(X_1 < 2x_2)}[/tex]
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)##.
 
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  • #5
rayge
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My guess is that it's the probability that [itex]X_1[/itex] is less than the mean of [itex]X_2[/itex], not some [itex]x_2[/itex].
 
  • #6
Ray Vickson
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My guess is that it's the probability that [itex]X_1[/itex] is less than the mean of [itex]X_2[/itex], not some [itex]x_2[/itex].

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.
 

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