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October 29th, 2017, 10:29 AM  #1 
Member Joined: Nov 2016 From: Kansas Posts: 65 Thanks: 0  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: Southern California, USA Posts: 1,604 Thanks: 817 
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 $E[e^X] = \dfrac{\alpha}{\alpha1}$ 
October 29th, 2017, 12:12 PM  #3  
Member Joined: Nov 2016 From: Kansas Posts: 65 Thanks: 0  Quote:
 
October 29th, 2017, 12:26 PM  #4  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,604 Thanks: 817  Quote:
$\displaystyle \int_0^\infty~\alpha e^{(1\alpha) t}~dt =$ $\left . \dfrac{\alpha}{1\alpha} e^{(1\alpha)t}\right _0^\infty = $ $\dfrac{\alpha}{\alpha  1}$  

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continuous, expected, random variable 
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