Finding probability given joint pdf.

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Homework Statement



Let X1 X2 X3 be independent and identically distributed random variables with common pdf f(x) = e^-x 0<x<infinity, zero elsewhere.

Evaluate P(X1 < X2|X1 < 2X2) and P(X1 < X2 < X3|X3 < 1).


The Attempt at a Solution



Would appreciate any hints as to where to start...
 
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I haven't actually studied any statistics so there might be an easier way of doing this but this is how I'd do it...

Having a conditional probability means that you essentially have a 100% probability for the condition, that is

\int_{x_1 &lt; 2x_2} dx_1 dx_2 f(x_1, x_2) = 1

This gives you the probability density given the condition that x_1 < 2 x_2. You can then use that to find the probability that x_1 < x_2.
 
clamtrox said:
I haven't actually studied any statistics so there might be an easier way of doing this but this is how I'd do it...

Having a conditional probability means that you essentially have a 100% probability for the condition, that is

\int_{x_1 &lt; 2x_2} dx_1 dx_2 f(x_1, x_2) = 1

This gives you the probability density given the condition that x_1 < 2 x_2. You can then use that to find the probability that x_1 < x_2.

But I don't think the problem is saying X1 < 2X2 I think the phrasing was X2|X1 < 2X2. Am I thinking about it wrong?
 
The first problem reads out as "probability that x_1 < x_2, given that you already know that x_1 < 2x_2". The 2nd problem is "probability that x_1 < x_2 < x_3, given that you know that x_3 < 1". Atleast that is what this notation means as far as I know. :-)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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