Chapter 7: Q 7.3 limits for convolution

Discussion in 'CT6' started by p_0910, Apr 4, 2017.

  1. p_0910

    p_0910 Keen member

    Question: If N takes values 0,1 and 2 with probabilities 0.5, 0.25 and 0.25 respectively, and the Xi's have a U(0,10) distribution, draw a sketch of the frequency distribution of S.

    After we obtain the density function for s in terms of an integrand involving the density function of X=s-y, how do we determine the limits for the integrand in terms of s?
    Thanks in advance.
     
  2. John Lee

    John Lee ActEd Tutor Staff Member

    I think you're confusing how the formulae work.

    \(P(S \leq x) = \sum\limits_{n=0}^{2} P(N=n)P(S \leq x | N=n)\)

    When N=0 then S=0.

    When N=1 then \(P(S \leq x | N=1) = P(X_1 \leq x)\)

    We can use integration or the CDF of U(0,10) to obtain this.

    When N=2 then \(P(S \leq x | N=2) = P(X_1 + X_2 \leq x)\)

    We'll need to use convolutions to obtain the answer to this. Setting \(Z = X_1 + X_2\) we have:

    \(P(Z \leq x) = \int\limits_0 ^x f_{X_1} (x_1) f_{X_2} (x-x_1) dx_1 \)
     

Share This Page