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#1
07-13-2012, 07:39 PM
 Bananas88 Member CAS SOA Join Date: May 2012 Location: NYC Studying for C/4 Posts: 137
TIA Exam 4 Question 3

I tried googling this problem, and searching on here for anything with a similar issue and I'm utterly lost.

The question says
Suppose that X is a Poisson random variable with mean 2.5 and Y is a geometric random variable on 1,2... with mean 2.5. Let P denote the probability that X is equal to its mode, and let Q denote the probability that Y is equal to its mode. Find |P-Q|.

My problem is with finding Q (I got the same P as the online solution did). For a geometric series, I thought that the mean is equal to (1-p)/p, so shouldn't p be equal to 2/7? In the solution they have the mean being 1/p=1/2.5, so, since the mode is 1, they got Q as being 0.4, while I had it as 2/7.

What am I doing wrong here?

Thanks for any help!
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#2
07-13-2012, 07:56 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,562

Quote:
 Originally Posted by Bananas88 Y is a geometric random variable on 1,2... with mean 2.5.
There are two versions of the geometric distribution. One on {1,2,...} the other on {0,1,2,...}. They told you to use the first. You used the second.
#3
07-13-2012, 07:58 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,562

See the thread containing this post

http://www.actuarialoutpost.com/actu...93#post4684993

or many other such threads found by searching for geometric
#4
07-13-2012, 08:28 PM
 Actuarialsuck Member Join Date: Sep 2007 Posts: 5,361

Or a wikipedia article on geometric distribution
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#5
07-13-2012, 09:13 PM
 Bananas88 Member CAS SOA Join Date: May 2012 Location: NYC Studying for C/4 Posts: 137

I didn't realize there was a difference in means for the two, thanks a ton for pointing it out/explaining, I really appreciate it!
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