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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: 2,585 Thanks: 1430 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: 3,264 Thanks: 902 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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 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. Tags distribution, expected, expected value, random variable Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ButterflyM Advanced Statistics 0 March 22nd, 2013 04:13 PM asd Advanced Statistics 5 March 19th, 2012 02:29 PM Anton29 Advanced Statistics 0 March 13th, 2012 09:59 PM pskye Algebra 0 April 4th, 2011 04:50 PM kahalla Advanced Statistics 0 March 9th, 2010 12:53 AM

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