Consider the process {X(t) , t >= 0} defined by X(t) = tY where Y is a uniformly distributed random variable on the interval (0,1). Give the expectation and the variance of X. Solution Y has the uniform distribution on (0,1), so E[Y] = (1-0)/2 = 1/2 Var[Y] = (1/12) (1-0)^2 = 1/12 Now X = ty (t positive) E[X] = E[tY] = t E[Y] = t/2 Var[X] = Var[tY] = t^2 Var[Y] = (t^2)/12.