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 mixture distribution $(9/10)N(0,1)+(1/10)N(0,9)$ is indeed $8.34$.

Step-by-step solution

Determine the Weights and Components

Here the two components of the mixture are $N(0,1)$ and $N(0,9)$ with weights $9/10$ and $1/10$ respectively. The first component is a standard normal distribution and the second is a normal distribution with variance 9.

Compute the Kurtosis of Each Component

Kurtosis of a standard normal distribution $N(0,1)$ is 3. Kurtosis of a normal distribution $N(0,9)$ is also 3.

Compute the Fourth Moment of Each Component

The fourth moment of a standard normal distribution $N(0,1)$ is 3. For the standard normal distribution with variance 9, we get the fourth moment by multiplying the fourth moment of the standard normal distribution by the fourth power of standard deviation which is $3\ast 81=243$.

Combine the Fourth Moments

A mixture's fourth moment is a weighted sum of each component's fourth moment. With the weights given $0.9$ for $N(0,1)$ and $0.1$ for $N(0,9)$, the fourth moment of the mixture is $0.9\ast 3+0.1\ast 243=27$.

Compute the Variance of the Mixture

The variance of the mixture is also the weighted sum of each component's variance and is given by $0.9\ast 1+0.1\ast 9=1.8$.

Compute the Kurtosis of the Mixture

Kurtosis of the mixture is the fourth moment divided by the square of the variance. Thus, the kurtosis of this mixture is $27/(1.8{)}^2=8.34$.