My Math Forum Expected Value

 October 29th, 2017, 10:29 AM #1 Member   Joined: Nov 2016 From: Kansas Posts: 73 Thanks: 1 Expected Value For $\alpha$ > 1, suppose that the continuous random variable X has the density function given by: f$_X$(t)= $\alpha$e$^\alpha$$^t$ if t $\geq$ 0 0 otherwise Find E[e$^X$] I'm getting the solution as infinity. Is it correct?
 October 29th, 2017, 11:24 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,317 Thanks: 1230 Yes, what you've written doesn't converge. I highly suspect the distribution is actually $f_X(t) = \alpha e^{-\alpha t}$ i.e. an exponential distribution. The function you gave is not a distribution. When the proper distribution is used $E[e^X] = \dfrac{\alpha}{\alpha-1}$ Thanks from ZMD
October 29th, 2017, 12:12 PM   #3
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Quote:
 Originally Posted by romsek Yes, what you've written doesn't converge. I highly suspect the distribution is actually $f_X(t) = \alpha e^{-\alpha t}$ i.e. an exponential distribution. The function you gave is not a distribution. When the proper distribution is used $E[e^X] = \dfrac{\alpha}{\alpha-1}$
For the distribution you stated, how did you end up with the given expected value?

October 29th, 2017, 12:26 PM   #4
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Quote:
 Originally Posted by ZMD For the distribution you stated, how did you end up with the given expected value?
$E[e^X] = \displaystyle \int_0^\infty~\alpha e^t e^{-\alpha t}~dt =$

$\displaystyle \int_0^\infty~\alpha e^{(1-\alpha) t}~dt =$

$\left . \dfrac{\alpha}{1-\alpha} e^{(1-\alpha)t}\right |_0^\infty =$

$\dfrac{\alpha}{\alpha - 1}$

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