Understanding Excess Kurtosis from Central Moments

This question asks us to calculate the excess kurtosis of a distribution given its second and fourth central moments. Kurtosis is a measure that describes the "tailedness" of a probability distribution's peak and tails relative to a normal distribution. Excess kurtosis is kurtosis minus 3.

What are Central Moments?

Central moments are specific types of moments of a probability distribution. The \(k\)-th central moment, denoted by \(\mu_k\), is the expected value of \((X - \mu)^k\), where \(X\) is the random variable and \(\mu\) is the mean of the distribution. Different central moments describe different characteristics of the distribution:

• The first central moment (\(\mu_1\)) is always 0.
• The second central moment (\(\mu_2\)) is the variance.
• The third central moment (\(\mu_3\)) measures skewness.
• The fourth central moment (\(\mu_4\)) measures kurtosis.

Calculating Kurtosis (\(\beta_2\))

Kurtosis is typically calculated using the formula involving the fourth and second central moments:

$$\beta_2 = \frac{\mu_4}{\mu_2^2}$$

We are given the following values:

• Second central moment, \(\mu_2 = 8\)
• Third central moment, \(\mu_3 = 50\) (Note: This is not needed for kurtosis calculation)
• Fourth central moment, \(\mu_4 = 199\)

Let's substitute the given values of \(\mu_4\) and \(\mu_2\) into the formula for kurtosis:

$$\beta_2 = \frac{199}{8^2} = \frac{199}{64}$$

Now, we calculate the value:

$$\beta_2 = \frac{199}{64} \approx 3.109375$$

Calculating Excess Kurtosis (\(\gamma_2\))

Excess kurtosis is defined as the kurtosis minus 3. This is done because a normal distribution has a kurtosis of exactly 3. Excess kurtosis tells us how much the distribution's kurtosis differs from that of a normal distribution.

The formula for excess kurtosis is:

$$\gamma_2 = \beta_2 - 3$$

Using the calculated value of \(\beta_2\):

$$\gamma_2 = \frac{199}{64} - 3$$

To subtract 3, we can express 3 with a denominator of 64: \(3 = \frac{3 \times 64}{64} = \frac{192}{64}\).

So, the excess kurtosis is:

$$\gamma_2 = \frac{199}{64} - \frac{192}{64} = \frac{199 - 192}{64} = \frac{7}{64}$$

Now, let's convert the fraction to a decimal:

$$\gamma_2 = \frac{7}{64} \approx 0.109375$$

Comparing with Options

The calculated excess kurtosis value is approximately 0.109375. Let's look at the given options:

• Option 1: 1.109
• Option 2: 0.109
• Option 3: 3.109
• Option 4: 2.109

Our calculated value, 0.109375, is closest to 0.109.

Summary of Calculation

Given:

• \(\mu_2 = 8\)
• \(\mu_4 = 199\)

Kurtosis \(\beta_2 = \frac{\mu_4}{\mu_2^2} = \frac{199}{8^2} = \frac{199}{64}\)

Excess Kurtosis \(\gamma_2 = \beta_2 - 3 = \frac{199}{64} - 3 = \frac{199 - 192}{64} = \frac{7}{64} \approx 0.109\)

Revision Table: Key Concepts

Additional Information on Kurtosis and Moments

Excess kurtosis helps us understand the shape of a probability distribution compared to a normal distribution:

• If excess kurtosis is close to 0, the distribution is mesokurtic (like a normal distribution).
• If excess kurtosis is positive (> 0), the distribution is leptokurtic. This means it has heavier tails and a sharper peak than a normal distribution.
• If excess kurtosis is negative (< 0), the distribution is platykurtic. This means it has lighter tails and a flatter peak than a normal distribution.

In this problem, the excess kurtosis is approximately 0.109, which is positive. This suggests the distribution is slightly leptokurtic, meaning it has slightly heavier tails than a normal distribution.

While the third central moment (\(\mu_3\)) was given, it relates to skewness, which measures the asymmetry of the distribution. A skewness of 0 indicates a perfectly symmetric distribution. Skewness is often calculated using the formula \(\gamma_1 = \frac{\mu_3}{\mu_2^{3/2}}\).