a) Var (X) =0
=> E (X − µ)2 = 0
=> X = µ ie. there is only one value that X takes ie. µ in this case. Hence, the distribution if degenerate since the variation among the variables is 0 which is only possible if the random variable takes only one value.
Now if X is a constant then the mean is also the same constant.
As a result X- µ = 0 , therefore as a resultant of this Var(X) = 0.
b) X : no.of heads - no.of tails = Y - Z (say)
Also Y+Z = n
Here, Y~binomial (n , p) and Z ~binomial( n, q)
E(X) = E( Y-Z) = E(Y) - E(Z) = np - nq = n(p-q)
Var(X) = Var( Y-Z) = Var(Y) + Var(Z) - 2cov (Y,Z) = npq + npq - 2cov( Y , n - Y) = npq+npq+ 2npq = 4npq
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