
Advanced Statistics Advanced Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
October 29th, 2017, 10:29 AM  #1 
Member Joined: Nov 2016 From: Kansas Posts: 68 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: USA Posts: 1,755 Thanks: 899 
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: 68 Thanks: 0  Quote:
 
October 29th, 2017, 12:26 PM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 1,755 Thanks: 899  Quote:
$\displaystyle \int_0^\infty~\alpha e^{(1\alpha) t}~dt =$ $\left . \dfrac{\alpha}{1\alpha} e^{(1\alpha)t}\right _0^\infty = $ $\dfrac{\alpha}{\alpha  1}$  

Tags 
continuous, expected, random variable 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Expected value of sde  amm  Advanced Statistics  1  June 26th, 2016 04:52 AM 
Expected Value  statssav  Probability and Statistics  5  June 11th, 2015 06:33 AM 
Expected Value  jbergin  Advanced Statistics  0  December 30th, 2014 10:29 AM 
p.m.f and expected value  baz  Algebra  4  September 11th, 2011 01:39 PM 
Expected value  kahalla  Advanced Statistics  0  March 9th, 2010 01:53 AM 