(New page: Starting this problem is similar to question 3 in homework 1. <math>Y = min(x_1, x_2)</math> <math>f(x_1) = c_1\cdot e^{-c_1x1}</math> <math>f(x_2) = c_2\cdot e^{-c_2x2}</math> From h...)
 
 
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I believe we can set the bounds from 0 to 1 since probabilities can not exceed this range.
 
I believe we can set the bounds from 0 to 1 since probabilities can not exceed this range.
  
<math>\int_0^1 \int_0^{x_2}  x_1 dx_1 dx_2 + \int_0^1 \int_{x_2}^1 x_2 dx_1 dx_2</math>
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<math>\int_0^1 \int_0^{f(x_2)f(x_1) dx_1 dx_2 + \int_0^1 \int_{f(x_2)}^1 f(x_2) dx_1 dx_2</math>
  
 
This equation is derived through the property that if <math>x_1</math> is smaller than <math>x_2</math> then <math>x_1</math> will be equal to Y and vise-versa
 
This equation is derived through the property that if <math>x_1</math> is smaller than <math>x_2</math> then <math>x_1</math> will be equal to Y and vise-versa
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I don't really see how the limit is f(x2) and you are also integrating on f(x2). Although I do agree the problem bears
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resemblance to hw 1, where we had to find the max of (x,y). Only in this case, it is an exponential random variable and you are
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finding the min. //Anand

Latest revision as of 15:27, 6 October 2008

Starting this problem is similar to question 3 in homework 1.

$ Y = min(x_1, x_2) $

$ f(x_1) = c_1\cdot e^{-c_1x1} $

$ f(x_2) = c_2\cdot e^{-c_2x2} $

From here we can use properties of integration to expand our min function into integrals.

$ \int_0^1 \int_0^1 min(x_1, x_2) $

I believe we can set the bounds from 0 to 1 since probabilities can not exceed this range.

$ \int_0^1 \int_0^{f(x_2)} f(x_1) dx_1 dx_2 + \int_0^1 \int_{f(x_2)}^1 f(x_2) dx_1 dx_2 $

This equation is derived through the property that if $ x_1 $ is smaller than $ x_2 $ then $ x_1 $ will be equal to Y and vise-versa


I don't really see how the limit is f(x2) and you are also integrating on f(x2). Although I do agree the problem bears resemblance to hw 1, where we had to find the max of (x,y). Only in this case, it is an exponential random variable and you are finding the min. //Anand

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