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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 | 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 find the max of (x,y). Only in this case, it is an exponential random variable and you are | + | 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 | finding the min. //Anand | ||
Latest revision as of 16:27, 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
