Results 1 to 5 of 5

Math Help - Variance of the sum of N independent variables

  1. #1
    Member
    Joined
    Jan 2009
    Posts
    83

    Variance of the sum of N independent variables

    Let X1,X2,...XN be independent identically distributed random variables, where N is a
    non-negative integer valued random variable. Let Z = X1 + X2 + : : + XN, (assuming that Z = 0 if N = 0). Find E(Z) and show that
    var(Z) = var(N)E(X1)^2 + E(N) var (X1)

    Not a clue on this one! I know Expectation of Z is NE(X1)

    but how do I show this result?

    Many Many thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by James0502 View Post
    Let X1,X2,...XN be independent identically distributed random variables, where N is a
    non-negative integer valued random variable. Let Z = X1 + X2 + : : + XN, (assuming that Z = 0 if N = 0). Find E(Z) and show that
    var(Z) = var(N)E(X1)^2 + E(N) var (X1)

    Not a clue on this one! I know Expectation of Z is NE(X1)

    but how do I show this result?

    Many Many thanks
    You're expected to know the theorem that for independent random variables Var\left[\sum_{i=1}^n (a_i X_i) \right]= \sum_{i=1}^n a_i^2 Var(X_i).

    Apply this theorem to your problem.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2008
    Posts
    5
    This is a problem of a random sum of random variables, which is very useful in actuarial science.
    First, I saw u already know E(sum Xi)= N*E(Xi) (A). keep this result, it's gotta be useful later.
    Second, remember var(sum Xi)=E(var(sum Xi|N=n))+var(E(sum Xi|N=n)) (B).

    We will work out the second part in the right side of equation (B);
    var(E(sum Xi|N=n))=var(E(sum Xi)) since Xi's and N are independent
    OK, substitute the E(sum Xi) result (A)in,
    and var(E(sum Xi|N=n))=var(N*E(Xi))=var(N)*(E(Xi))^2
    hew...

    Here come the first part in the right side of equation (B),
    E(var(sum Xi|N=n))=E(N*var(X)) since Xi's are independent and X and N are indep.
    then, E(var(sum Xi|N=n))=E(N)*var(X)

    Combine two parts, u get the result u expected. u need to show a bit more detail when u prove the first part in equation (B). I skipped some steps. If you want to be an actuary, we should talk more!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jan 2009
    Posts
    83
    many thanks guys!

    I am still unsure how to get (B) though. I cant think of any useful identities I know to give me this result - we've only really just begun this section of the course, so I'm still very hazy!

    thanks
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    Quote Originally Posted by James0502 View Post
    Let X1,X2,...XN be independent identically distributed random variables, where N is a
    non-negative integer valued random variable. Let Z = X1 + X2 + : : + XN, (assuming that Z = 0 if N = 0). Find E(Z) and show that
    var(Z) = var(N)E(X1)^2 + E(N) var (X1)

    Not a clue on this one! I know Expectation of Z is NE(X1)

    but how do I show this result?

    Many Many thanks
    N is a random variable, known as a stopping time.
    So E(Z)=NE(X_1) does not make sense, the left side is constant while the right is random.
    I don't know if we need independence or not between the N and the
    X_i's in the mean case, but we may need it, in obtaining the variance.
    Most likely E(Z)=E(N)E(X_1).

    I just looked in up in my advisor's advisor's book.
    This is correct, it's known as Wald's Equation.
    proof is in...
    http://en.wikipedia.org/wiki/Wald's_equation

    I also found the proof to the variance on page 139 of that book.
    Last edited by matheagle; February 24th 2009 at 05:41 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Independent Random Variables - Variance of 2 Die
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 29th 2010, 10:54 PM
  2. independent variables
    Posted in the Advanced Statistics Forum
    Replies: 7
    Last Post: November 20th 2010, 12:19 AM
  3. Variance of the product of two independent variables
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: December 17th 2009, 12:20 PM
  4. Replies: 3
    Last Post: December 14th 2009, 08:03 PM
  5. Independent variables
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: January 24th 2009, 10:41 AM

Search Tags


/mathhelpforum @mathhelpforum