Results 1 to 5 of 5

Math Help - Conditional expectation of a PMF

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    11

    Conditional expectation of a PMF

    I'm having trouble starting this problem.

    Say you committed a crime and you're sentenced to jail. When entering the jail, you pick one ball from a box containing 3 balls, each numbered 0, 1, and 3. If you select 0, you get out of jail; if you select 1 or 3, you put the ball back into the box and after that number of years, you select again under the same conditions. You repeat this until you select a 0 and you're free to go. What is the expected value of how long you'll be in jail for?

    The hint is that you're suppose to condition on the first ball selected. I don't see how this helps?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member TriKri's Avatar
    Joined
    Nov 2006
    Posts
    358
    Thanks
    1
    Set E as the expected time to be in jail. Now there is three cases:

    1 (1/3 probability): 0 years
    2 (1/3 probability): 1 year, then you will have to do the same thing again, so in average 1 + E
    3 (1/3 probability): 3 years, then you will have to do the same thing again, so in average 3 + E

    To add it up, E = (1/3)*0 + (1/3)*(1+E) + (1/3)*(3+E). Now this is a regular linear equation.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2009
    Posts
    11
    Quote Originally Posted by TriKri View Post
    Set E as the expected time to be in jail. Now there is three cases:

    1 (1/3 probability): 0 years
    2 (1/3 probability): 1 year, then you will have to do the same thing again, so in average 1 + E
    3 (1/3 probability): 3 years, then you will have to do the same thing again, so in average 3 + E

    To add it up, E = (1/3)*0 + (1/3)*(1+E) + (1/3)*(3+E). Now this is a regular linear equation.
    Solving that equation for E, I get E = 11 years.

    I don't understand why for 1 year, the average would be 1 + E. Is it because after 1 year, you select again and the expectation of selecting then is the same as selecting now?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2009
    Posts
    1
    I think E = 4, try to resolve the linear equation again.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member TriKri's Avatar
    Joined
    Nov 2006
    Posts
    358
    Thanks
    1
    Quote Originally Posted by CUEngineering View Post
    Solving that equation for E, I get E = 11 years.
    E = 11 doesn't solve the equation. Try to solve it again; it should be 4, as impossibletask wrote.

    Quote Originally Posted by CUEngineering View Post
    I don't understand why for 1 year, the average would be 1 + E. Is it because after 1 year, you select again and the expectation of selecting then is the same as selecting now?
    Yes, that's exactly why.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: July 22nd 2011, 01:39 AM
  2. Conditional Expectation
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: March 29th 2011, 05:01 PM
  3. Conditional Expectation
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: January 29th 2010, 08:34 PM
  4. Expectation & Conditional Expectation
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: February 1st 2009, 10:42 AM
  5. Conditional Expectation
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: December 9th 2007, 08:22 PM

Search Tags


/mathhelpforum @mathhelpforum