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December 28th, 2016, 04:21 PM  #1 
Newbie Joined: Nov 2016 From: Israel Posts: 7 Thanks: 0  Expectation calculation
Hi, I could only find E(XY1=1) is 7, but had difficulties with all the rest. I would be glad for a help. Thanks in advance.

December 28th, 2016, 04:25 PM  #2  
Newbie Joined: Nov 2016 From: Israel Posts: 7 Thanks: 0  Quote:
Last edited by skipjack; December 29th, 2016 at 08:35 AM.  
December 28th, 2016, 06:52 PM  #3 
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 763 Thanks: 278 Math Focus: Linear algebra, linear statistical models 
Since the monkey is right back at the beginning of the maze after going through the other two apertures, E(XY1=2) = 3 + E(X), E(XY1=2) = 5 + E(X). We have $\displaystyle E(X) = \sum_{i=1}^3 E(XY1=i) P(Y1=i) = \frac{1}{3} [7 + 3 + E(X) + 5 + E(X)]$, so you simply have to solve for E(X) to get the answer! 
December 29th, 2016, 06:15 AM  #4 
Newbie Joined: Nov 2016 From: Israel Posts: 7 Thanks: 0 
Hi 123qwerty, first of all, thanks for your reply. Can you please explain the equations E(XY1=2) = 3 + E(X), E(XY1=2) = 5 + E(X)? I mean, I undestand that the 3 and the 5 are the periods of time by which the monkey's moving about through each aperture except 1, but I don't undestand what makes the the 2 equations true. And I would be greatful if you'd help me with the second part of section b. Last edited by riovelo; December 29th, 2016 at 06:26 AM. 
December 29th, 2016, 07:48 AM  #5 
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 763 Thanks: 278 Math Focus: Linear algebra, linear statistical models 
We're splitting up the expected value into two parts: the first time period taken by the monkey to move about in the maze, plus everything else. The 'everything else' part is E(X) because we're back to the beginning of the maze and have to start over. I'm not sure what the last one really means tbh, but here's how I'd do it: We know for certain that he'll go through 7 once, so that's the first term. As for how much time (on average) he spend on going through the rest of the maze, well, let this time be Z. We know he'll choose 3 half the time and 5 the other half of the time, so let's just pretend there's a 2/3 chance he'll choose a path with 4 hours. So now we have: Then we get the sum $\displaystyle 7 + \sum_{i=1}^\infty \left(\frac{2}{3}\right)^{i1} \left(\frac{1}{3}\right) \times 4(i1) = 7 + 2 \times 4 = 15$ (Note that I put i1 rather than i after 4 to exclude the first round.) 
December 29th, 2016, 09:53 AM  #6 
Newbie Joined: Nov 2016 From: Israel Posts: 7 Thanks: 0 
Thanks again 123qwerty. But stil, there's one thing I don't understand Why the first time period taken by the monkey is not included in that 'everything else'?

December 29th, 2016, 09:57 AM  #7 
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 763 Thanks: 278 Math Focus: Linear algebra, linear statistical models  
December 29th, 2016, 07:17 PM  #8  
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 763 Thanks: 278 Math Focus: Linear algebra, linear statistical models  Quote:
 
December 30th, 2016, 03:44 AM  #9 
Newbie Joined: Nov 2016 From: Israel Posts: 7 Thanks: 0  

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calculation, expectation, expectaton, probabability 
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