Solution to Exercise 4.24.

  (a)

$$E[X] \; = \; \int_{\theta}^{\infty} \, x \frac{k \theta^k}{x^{k+1}} \, dx$$

$$\; = \; k \theta^k \int_{\theta}^{\infty} \, x^{-k} \, dx \; = \; \frac{k}{k-1} \theta .$$

(b) If $k \le 1$, then the integral defining the mean is $\infty$.

(c)

$$E[X^2] \; = \; k \theta^k \int_{\theta}^{\infty} \, x^{-(k-1)} \, dx \; = \; \frac{k}{k-2} \theta^2 .$$

Using the ``short cut'' formula for variance on p. 156,

$$V[X] \; = \; \frac{k}{k-2} \theta^2 - \left( \frac{k}{k-1} \... ...\frac{k}{k-2} - \left( \frac{k}{k-1} \right)^2 \right] \theta^2$$

$$\; = \; \frac{k(k-1)^2 - k^2(k-2)}{(k-2)(k-1)^2} \theta^2 \... ...(k-2)(k-1)^2} \theta^2 \; = \; \frac{k}{(k-2)(k-1)^2} \theta^2$$

(d) If $k \le 2$, then the integral defining E[X²] is $\infty$.Since E[X] is OK when k=2, we could say $V[X] = \infty$ when k=2 (actually, when $1 < k \le 2$).

(e)

$$\int_{\theta}^{\infty} \, x^{n-k-1} \, dx \; < \; \infty \mbox{ if and only if } n-k-1 \gt -1 , \mbox{ i.e. } n \gt k .$$