∈ {0, 1}K , which a particular element zk is equal to 1 and all other elements equal to 0. There are K possible outcomes for this binary vector. Let πk = p(zk = 1) (i.e., the mixing coefficient), and recall that πk is a pmf (i.e., πk ∈ [0, 1] and πk = 1). The probability of z is, p(z) = p(z1 , . . . , zk ) = K k=1 πzk k Similarly, the conditional distribution of x given a particular value for z is a Gaussian – p(x|zk = 1) = N(x|µk , Σk ). Then, p(x|z) = p(x|z1 , . . . , zk ) = K k=1 N(x|µk , Σk )zk Marginalizing p(x) yeilds p(x) = z∈Z p(x|z)p(z) = K k=1 πk N(x|µk , Σk ) Well isn’t that interesting! The marginal on x can be determined using a set of latent variables. Generally it is easier to work with p(x, z) over p(x). 62 / 69