Question

Let the random variable \(X\) have pmf $$ p(x)=\left\\{\begin{array}{ll} p & x=-1,1 \\ 1-2 p & x=0 \\ 0 & \text { elsewhere } \end{array}\right. $$ where \(0

Short answer

The kurtosis for different \(p\) values are as follows: for \(p = \frac{1}{3}\), kurtosis is 1.5; for \(p = \frac{1}{5}\), kurtosis is 2.5; for \(p = \frac{1}{10}\), kurtosis is 5; and for \(p = \frac{1}{100}\), kurtosis is 50.

Step-by-step solution

Compute the fourth moment

The fourth moment (about the origin) of a random variable \(X\) is given by \(E[X^4] = \sum(p(x).x^4)\). In this case, \(E[X^4] = (1 * (-1)^4 * p) + (1 * (1)^4 * p) + (0 * 0^4 * (1-2p)) = 2p.

Compute the second moment

The second moment (about the origin) of a random variable \(X\) is given by \(E[X^2] = \sum(p(x).x^2)\). In this case, \(E[X^2] = (1 * (-1)^2 * p) + (1 * (1)^2 * p) + (0 * 0^2 * (1-2p)) = 2p.

Compute the kurtosis

Kurtosis is given by the ratio of the fourth moment to the square of the second moment, i.e., \(Kurtosis = E[X^4] / (E[X^2])^2\). Substituting the values we obtained in steps 1 and 2, we get \(Kurtosis = 2p / (2p)^2 = 1 / (2p)\).

Substitute the given values of p

Substitute the given values of \(p = \frac{1}{3}\), \(p = \frac{1}{5}\), \(p = \frac{1}{10}\), and \(p = \frac{1}{100}\) into the kurtosis formula to get respective values of kurtosis. For \(p = \frac{1}{3}\), \(Kurtosis = \frac{1}{2*\frac{1}{3}} = 1.5\). For \(p = \frac{1}{5}\), \(Kurtosis = \frac{1}{2*\frac{1}{5}} = 2.5\). For \(p = \frac{1}{10}\), \(Kurtosis = \frac{1}{2*\frac{1}{10}} = 5\). Finally, for \(p = \frac{1}{100}\), \(Kurtosis = \frac{1}{2*\frac{1}{100}} = 50\). As we can see, the kurtosis increases as \(p\) decreases, which verifies the statement in the question.