(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...) |
|||
| Line 13: | Line 13: | ||
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> | + | <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 | ||
Revision as of 10:21, 5 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
