My Math Forum  

Go Back   My Math Forum > High School Math Forum > Probability and Statistics

Probability and Statistics Basic Probability and Statistics Math Forum


Thanks Tree5Thanks
  • 2 Post By 123qwerty
  • 1 Post By 123qwerty
  • 1 Post By 123qwerty
  • 1 Post By 123qwerty
Reply
 
LinkBack Thread Tools Display Modes
December 28th, 2016, 03:21 PM   #1
Newbie
 
Joined: Nov 2016
From: Israel

Posts: 7
Thanks: 0

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.
Attached Images
File Type: jpg probability.jpg (21.9 KB, 15 views)
riovelo is offline  
 
December 28th, 2016, 03:25 PM   #2
Newbie
 
Joined: Nov 2016
From: Israel

Posts: 7
Thanks: 0

Quote:
Originally Posted by riovelo View Post
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 07:35 AM.
riovelo is offline  
December 28th, 2016, 05:52 PM   #3
Senior Member
 
Joined: Dec 2012
From: Hong Kong

Posts: 853
Thanks: 311

Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics
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
123qwerty is offline  
December 29th, 2016, 05:15 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 05:26 AM.
riovelo is offline  
December 29th, 2016, 06:48 AM   #5
Senior Member
 
Joined: Dec 2012
From: Hong Kong

Posts: 853
Thanks: 311

Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics
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
123qwerty is offline  
December 29th, 2016, 08:53 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'?
riovelo is offline  
December 29th, 2016, 08:57 AM   #7
Senior Member
 
Joined: Dec 2012
From: Hong Kong

Posts: 853
Thanks: 311

Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics
Quote:
Originally Posted by riovelo View Post
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.
Thanks from riovelo
123qwerty is offline  
December 29th, 2016, 06:17 PM   #8
Senior Member
 
Joined: Dec 2012
From: Hong Kong

Posts: 853
Thanks: 311

Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics
Quote:
Originally Posted by 123qwerty View Post
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.
Thanks from riovelo
123qwerty is offline  
December 30th, 2016, 02:44 AM   #9
Newbie
 
Joined: Nov 2016
From: Israel

Posts: 7
Thanks: 0

Quote:
Originally Posted by 123qwerty View Post
Oops! Just realised this.
Thank you very much!
riovelo is offline  
Reply

  My Math Forum > High School Math Forum > Probability and Statistics

Tags
calculation, expectation, expectaton, probabability



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Expectation Francobati Advanced Statistics 4 February 2nd, 2016 12:45 PM
Integral calculation based on coordinates / Area under Curve calculation joskevermeulen Calculus 1 December 29th, 2015 05:02 AM
Expectation of (aX-bY) ? nikozm Algebra 3 December 18th, 2013 01:07 PM
proving expectation. asoracc Advanced Statistics 3 January 24th, 2011 03:49 PM
What is the expectation of this RV? STV Advanced Statistics 9 July 14th, 2008 11:45 AM





Copyright © 2017 My Math Forum. All rights reserved.