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October 29th, 2017, 10:29 AM   #1
ZMD
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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?
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October 29th, 2017, 11:24 AM   #2
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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

$E[e^X] = \dfrac{\alpha}{\alpha-1}$
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October 29th, 2017, 12:12 PM   #3
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Quote:
Originally Posted by romsek View Post
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?
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October 29th, 2017, 12:26 PM   #4
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Quote:
Originally Posted by ZMD View Post
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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