Supplement to handout on examples of convergence modes ------------------------------------------------------ In all the example, S=[0,1], Q=Borel([0,1]), P=uniform measure 1. For this example, take X to be the 0 function (i.e., X(z)=0 for all z in [0,1]). We will show that there is m.s. convergence (to X) but not a.s. convergence (m.s. immediately implies in probability). The m.s. value in this example is given by E(X_n-X)^2 = E(X_n)^2 = EX_n (because X_n take only the values 0 and 1) = area under the pulse (because P is the uniform measure). We can see that the areas under the pulses are going to 0. Therefore we have m.s. convergence. On the other hand, take any z in [0,1]. The sequence {X_n(z)} does NOT converge to zero because the sequence takes the value 1 infinitely often (and 0 otherwise). For example, if we look at z=0, the sequence {X_n(0)} is of the form 1,0,1,0,0,1,0,0,0,1... Therefore, the set of z in [0,1] for which the sequence {X_n(z)} converges to X (the zero function) is the empty set. Hence we do not have a.s. convergence. 2. For this example, again take X to be the 0 function. We will show that there is a.s. convergence (to X) but not m.s. convergence (a.s. immediately implies in probability). The m.s. value in this example is given by E(X_n-X)^2 = E(X_n)^2 = n*(1/n) = 1 Therefore, the m.s. value does not converge to 0, and hence we don't have m.s. convergence. On the other hand, for any z in (0,1], {X_n(z)} converges to 0. For example, if we look at z=1/3, the sequence {X_n(1/3)} is of the form 1, sqrt{2}, sqrt{3}, 0, 0, 0, ... Therefore, the set of z in [0,1] for which the sequence {X_n(z)} converges to X (the zero function) is the set (0,1], which has probability 1. Hence we have a.s. convergence. 3. In this example, the sequence {X_n} switches between the two functions shown (left function for odd values of n, right function for even n). The m.s. value E(X_n-X)^2 for both odd and even n are the same: 1/2. So the m.s. value does not converge to 0 (hence we don't have m.s. convergence). Moreover, for any z in [0,1], the sequence {X_n(z)} does not converge to 0. For example. {X_n(1/4)}={1,0,1,0,1,0,1...}. Hence we don't have a.s. convergence. On the other hand, both left and right random variables have exactly the same distribution: both are Bernoulli with P{X_n=1}=1/2 and P{X_n=0}=1/2. Therefore, the distribution of X_n converges (actually, remains constant over n, which is a degenerate special case). Hence, we have convergence in distribution.