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 October 6th, 2017, 04:14 PM #1 Member   Joined: Nov 2016 From: Kansas Posts: 73 Thanks: 1 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, 04:43 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,124 Thanks: 1102 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. $E[Y] = \displaystyle \int_e^{e^2}~y\dfrac 1 y~dy = \int_e^{e^2}~1=e^2-e$
 October 8th, 2017, 03:46 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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, 09:40 AM   #4
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Quote:
 Originally Posted by Country Boy 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.
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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