Distributions may in general be skewed, but there may be conditions on their parameters that make the skewness get smaller or even disappear. this problem uses moment-generating functions (mgfs) to explore that idea for two important discrete distributions, the binomial and the poisson. (a) we saw in class that if x ~ binomial (n, p), for 0 < p < 1 and integer n greaterthanorequalto 1, then the mgf of x is given by phi x (t) = [pe^t + (1 - p)]^n. for all real t, and we used this to work out the first three moments of x (note that the expression for e (x^3) is only correct for n greaterthanorequalto 3): e(x) = np. e (x^2) = np[(1 + (n - 1)p], e (x^3) = np[1 + (n - 2)(n - 1)p^2 + 3 (n - 1)p], from which we also found that v(x) = np(1 - p). show that the above facts imply that skewness (x) = 1-2 p/squareroot n p(1 - p). under what condition on p, if any, does the skewness vanish? under what condition on n, if any, does the skewness tend to 0? explain briefly. (b) in our brief discussion of stochastic processes we encountered the poisson distribution: if y ~ poisson (lambda). for lambda > 0. then the pf of y is fy(y) = {lambda^y e^- lambda/y! for y = 0, 1, 0 otherwise}. (i) use this to show that for all real t the mgf of y is psi_y (t) = e^lambda (e^t - 1). (ii) use psi_y(t) to compute the first three moments of y. the variance of y and the skewness of y. under what condition on a, if any, does the skewness either disappear or tend to 0? explain briefly.