Question

Consider the mixture distribution \((9 / 10) N(0,1)+(1 / 10) N(0,9) .\) Show that its kurtosis is \(8.34\).

Short answer

The kurtosis of the given mixture distribution is indeed 8.34.

Step-by-step solution

Calculate the fourth moment for each normal distribution

For a standard normal distribution with mean µ and variance σ^2, its fourth moment about the mean is given by E[(X-µ)^4] = 3 * σ^4. In this case, for the first distribution N(0,1), since σ=1, the fourth moment is 3*(1^4) = 3. For the second distribution N(0,9), the fourth moment is 3*(√9^4) = 3 * 81 = 243.

Calculate the weighted fourth moment for the mixture distribution

The fourth moment of the mixture distribution is given by the weights of the two normals times their respective fourth moments. Therefore, it is equal to (9/10)*3 + (1/10)*243 = 2.7 + 24.3 = 27.

Calculate the variance of the mixture distribution

Like the fourth moment, the variance of the mixture distribution is given by the weights of the two normals times their respective variances. Therefore, it is equal to (9/10)*1 + (1/10)*9 = 0.9 + 0.9 = 1.8.

Calculate the kurtosis of the mixture distribution

The kurtosis of a distribution is given by the fourth moment divided by the square of the variance. Therefore, the kurtosis of the mixture distribution is equal to 27/(1.8^2) = 8.34.