Results 1 to 3 of 3

Math Help - trinomial distribution

  1. #1
    Banned
    Joined
    Feb 2009
    Posts
    97

    trinomial distribution

    Let X and Y have a trinomial distribution with parameters n=3, p_{1}=\frac{1}{6} and p_{2}=\frac{1}{2}. Find:
    a) E(X)
    b) E(Y)
    c) Var(X)
    d) Var(Y)
    e) Cov(X,Y)
    f) \rho

    *Note that \rho=-\sqrt{p_{1}p_{2}/(1-p_{1})(1-p_{2})}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by antman View Post
    Let X and Y have a trinomial distribution with parameters n=3, p_{1}=\frac{1}{6} and p_{2}=\frac{1}{2}. Find:
    a) E(X)
    E(X)=np_1=3\left(\frac{1}{6}\right)=\boxed{\frac{1  }{2}}

    b) E(Y)
    Similarly, E(Y)=np_2=3\left(\frac{1}{2}\right)=\boxed{\frac{3  }{2}}

    c) Var(X)
    \text{Var}(x)=np_1(1-p_1)=3\left(\frac{1}{6}\right)\left(\frac{5}{6}\ri  ght)=\boxed{\frac{5}{12}}

    d) Var(Y)
    Similarly, \text{Var}(Y)=np_2(1-p_2)=3\left(\frac{1}{2}\right)\left(\frac{1}{2}\ri  ght)=\boxed{\frac{3}{4}}

    e) Cov(X,Y)
    I'm not too sure about the Covariance. I'm tempted to say that its -np_1p_2=-3\left(\frac{1}{6}\right)\left(\frac{1}{2}\right)=  \boxed{-\frac{1}{4}} (Someone verify this... -_-)


    f) \rho

    *Note that \rho=-\sqrt{p_{1}p_{2}/(1-p_{1})(1-p_{2})}
    You're given enough information to find \rho. Its just plug-n-chug from here...

    Does this make sense?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Chris L T521 View Post
    I'm not too sure about the Covariance. I'm tempted to say that its -np_1p_2=-3\left(\frac{1}{6}\right)\left(\frac{1}{2}\right)=  \boxed{-\frac{1}{4}} (Someone verify this... -_-)
    For one trial, \mathbb{E}(X_1Y_1)=\mathbb{P}(X_1=1,Y_1=1)=0, because both can't happen at the same time.

    \text{cov}(X_1,Y_1)=\mathbb{E}(X_1Y_1)-\mathbb{E}(X_1)\mathbb{E}(Y_1)=-p_1p_2

    Then, since in the trinomial distribution we have n independent trials, \text{cov}(X,Y)=-n\text{cov}(X_1,Y_1)=-np_1p_2
    ( X=\sum_{i=1}^n X_i and Y=\sum_{i=1}^n Y_i)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. DE trinomial?
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: August 30th 2011, 05:35 PM
  2. trinomial
    Posted in the Algebra Forum
    Replies: 2
    Last Post: December 8th 2009, 10:09 AM
  3. Little Bit o' Trinomial Help.
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 19th 2009, 09:04 AM
  4. Trinomial Distribution
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 23rd 2008, 09:29 PM
  5. trinomial
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 21st 2008, 06:55 PM

Search Tags


/mathhelpforum @mathhelpforum