Results 1 to 6 of 6

Math Help - conditioning problem - homework due very soon

  1. #1
    Member
    Joined
    Mar 2008
    Posts
    78

    conditioning problem - homework due very soon

    Suppose X|Y ~ poisson (mean = y) and Y ~ Uniform [0, 5]

    compute Pr [X >= 1]

    I think I did it, but need confirmation it's right

    E[X] = 2.5

    Pr [X >= 1 ] = 1 - e^(-2.5)

    this seems overly simple, but is it correct?
    Last edited by paulrb; March 3rd 2010 at 11:25 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by paulrb View Post
    Suppose X|Y ~ poisson (mean = y) and Y ~ Uniform [0, 5]

    compute Pr [X >= 1]

    I think I did it, but need confirmation it's right

    E[X] = 2.5

    Pr [X >= 1 ] = 1 - e^(-2.5)

    this seems overly simple, but is it correct?
    Pr(X\ge 1)=1-Pr(X=0)=1-\int_{\text{support}(y)} Pr(X=0|y)p(y)\;dy= 1-\int_{y=0}^5 \frac{Pr(X=0|y)}{5}\;dy =1-\frac{1}{5}\int_{y=0}^5  e^{-y} \;dy

    CB
    Last edited by CaptainBlack; March 5th 2010 at 02:56 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    But f_Y(y)=1/5 on 0<y<5

    Y\sim U(0,5) not U(0,1)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by matheagle View Post
    But f_Y(y)=1/5 on 0<y<5

    Y\sim U(0,5) not U(0,1)
    Oppss ... working on auto-pilot there.

    CB
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2010
    From
    TAIWAN
    Posts
    39
    Quote Originally Posted by CaptainBlack View Post
    Pr(X\ge 1)=1-Pr(X=0)=1-\int_{\text{support}(y)} Pr(X=0|y)p(y)\;dy= 1-\int_{y=0}^5 \frac{Pr(X=0|y)}{5}\;dy =1-\frac{1}{5}\int_{y=0}^5 e^{-y} \;dy

    CB
    can this be more simple?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by chialin4 View Post
    can this be more simple?
    The last integral is about as simple as you are going to get.

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Conditioning Number
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: November 15th 2011, 10:55 AM
  2. Conditioning
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: September 7th 2010, 10:01 PM
  3. Conditioning on a RV
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: February 1st 2010, 06:05 AM
  4. Conditioning problem
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: March 23rd 2009, 03:49 PM
  5. Serious help - Conditioning problem
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: March 23rd 2009, 03:34 PM

Search Tags


/mathhelpforum @mathhelpforum