My Math Forum Expectation calculation

 Probability and Statistics Basic Probability and Statistics Math Forum

December 28th, 2016, 04:21 PM   #1
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Expectation calculation

Hi, I could only find E(X|Y1=1) is 7, but had difficulties with all the rest. I would be glad for a help. Thanks in advance.
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December 28th, 2016, 04:25 PM   #2
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Quote:
 Originally Posted by riovelo Hi, I could only find E(X|Y1=1) is 7, but had difficulties with all the rest. I would be glad for a help. Thanks in advance.
There is the bigger image. http://up416.siz.co.il/up1/gmvix3jyzmwz.png

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: 781 Thanks: 284 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(X|Y1=2) = 3 + E(X), E(X|Y1=2) = 5 + E(X). We have $\displaystyle E(X) = \sum_{i=1}^3 E(X|Y1=i) P(Y1=i) = \frac{1}{3} [7 + 3 + E(X) + 5 + E(X)]$, so you simply have to solve for E(X) to get the answer! Thanks from romsek and riovelo
 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(X|Y1=2) = 3 + E(X), E(X|Y1=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: 781 Thanks: 284 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 $\displaystyle 7 + \sum_{i=1}^\infty \left(\frac{2}{3}\right)^{i-1} \left(\frac{1}{3}\right) \times 4(i-1) = 7 + 2 \times 4 = 15$ (Note that I put i-1 rather than i after 4 to exclude the first round.) Thanks from riovelo
 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
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Quote:
 Originally Posted by riovelo 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'?
Because I divided the time into two parts: The first part plus the rest.

December 29th, 2016, 07:17 PM   #8
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Quote:
 Originally Posted by 123qwerty 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 $\displaystyle 7 + \sum_{i=1}^\infty \left(\frac{2}{3}\right)^{i-1} \left(\frac{1}{3}\right) \times 4(i-1) = 7 + 2 \times 4 = 15$ (Note that I put i-1 rather than i after 4 to exclude the final round.)
Oops! Just realised this.

December 30th, 2016, 03:44 AM   #9
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Quote:
 Originally Posted by 123qwerty Oops! Just realised this.
Thank you very much!

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