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October 6th, 2017, 05:14 PM  #1 
Member Joined: Nov 2016 From: Kansas Posts: 68 Thanks: 0  Expected Value
X is a random variable over [1,2] 1. Find distribution function of Y=e$^x$ 2. Find E[Y] i.e expected value of Y I'm done with part a. The answer is 1/y . Can anyone help calculate the expected value. 
October 6th, 2017, 05:43 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,656 Thanks: 842 
The answer to part a is not $\dfrac 1 y$ It is $f_Y(y) = \begin{cases}0 & y < e \\ \dfrac 1 y &y \in [e,e^2] \\ 0 &e^2 < y \end{cases}$ There is a serious difference. $E[Y] = \displaystyle \int_e^{e^2}~y\dfrac 1 y~dy = \int_e^{e^2}~1=e^2e$ 
October 8th, 2017, 04:46 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,876 Thanks: 766 
Romsek is assuming you meant "x is uniformly distributed over [1, 2]". Just saying "x is a random variable over [1, 2]" does not tell us the probability distribution which is necessary in order to answer this question.

October 8th, 2017, 10:40 AM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 1,656 Thanks: 842  Yeah. It's fairly a given that if they don't mention some other distribution they mean uniform if it's over a closed interval.


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distribution, expected, expected value, random variable 
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