In general, consider a discrete random variable with probability massfunction f which gives the outcome a with probability P₁, i = 1,2,...,n.Suppose we want to create a new random variable y with probability massfunction g over the outcomes a which is a mean-preserving spread of f.We can create this new random variable y by adding uncorrelated "noise"to x in a manner that and y have the same mean. In particular, we canlet g give the outcome b; with probability Pi, i = 1, 2,...,n, where bi iseither the amount a; for sure or a lottery with expected value equal to ai.Then g will be a mean-preserving spread of f. You will now prove this forthe following probability mass function for a discrete random variable *:xa1 az a3 04ƒ(x)Assume that a₁ < A₂ < az < A4 and:Qg =a1 + as2a3 =a₂ + as2Construct a random variable y with probability mass function 9:y51 | 52 | 53 | 54g(y)H=where by: a₁, b = a4, b₂ is a simple lottery which gives either a₁ or aswith equal probability, and be is a simple lottery which gives either az ora₂ with equal probability. Show that y has the same mean as x but shiftsthe probability mass from the center of f towards the tails.

PMF f of discrete random variable x:xa1a2a3a4f(x)1/41/41/41/4Assume PMF g of discrete random variable y:yb1b2b3a4g(y)1/41/41/41/4Also given, b1=a1, b4=a4 , , b2 gives a1 and a3 with equal probability and b3 gives a2 and a4 with equal probability.Let's computed Expected value of X.Let's compute the mean of Y.Hence proved E(X) = E(Y) This proves that y has the same mean as x. The probabilities assigned to b2 and b3 are split over two values as b2 is a simple lottery of a1 and a3 and b3 is a simple lottery of a2 and a4. This means that y assigns more probability mass to its tails while x assigns equal probability for all values. This clearly illustrates that y is a mean preserving distribution that spreads the probability mass from the center towards its tails.This makes g a mean-preserving spread of f.We proved that y has the same mean as x so mean is preserved in y. However, probabilities assigned to b2 and b3 are split over two values. This means that y assigns more probability to its tails and spreads the probability mass from the center towards its tails.