6
$\begingroup$

Given the distribution funciton of the r.v. $X$ for $\alpha, \beta >0$

$$ F(x) = 1-\Big( \frac{\beta}{\beta +x}\Big)^{\alpha} $$

for $x \geq 0$ and $0$ elsewhere

What is the expectation and variance of $X$ for those values of parameters, where it is defined? Furthermore, how can one estimate the parameters $(\alpha, \beta)$ with given data using the method of moments?

My idea was to first of all calculate the density, i.e.

$$f(x) = \frac{\alpha \beta^{\alpha}}{(\beta + x)^{\alpha+1}}$$

for $x \geq 0$.

Then

$$E[X] = \alpha \beta^{\alpha}\int_0^{\infty}\frac{x}{(\beta + x)^{\alpha+1}} dx$$

How can I proceed now?

$\endgroup$
6
$\begingroup$

Use the substitution $u = x+\beta$, $x = u-\beta$, $dx = du$ to obtain $$\int_{u=\beta}^\infty (u-\beta) u^{-\alpha} \, du = \int_{u=\beta}^\infty u^{-\alpha+1} - \beta u^{-\alpha} \, du,$$ and continue from there.


A complete solution follows:

Differentiating the CDF gives the density $$f_X(x) = \frac{\alpha \beta^\alpha}{(\beta+x)^{\alpha+1}}, \quad x \ge 0.$$ Then consider the $k^{\rm th}$ non-central moment of $X$ about $-\beta$; i.e., $$\operatorname{E}[(X+\beta)^k] = \int_{x=0}^\infty (\beta+x)^k f_X(x) \, dx = \int_{x=0}^\infty \frac{\alpha \beta^\alpha}{(\beta+x)^{\alpha+1-k}} \, dx.$$ This of course is easily integrable using traditional methods: we find $$\operatorname{E}[(X+\beta)^k] = \left[\frac{-\alpha\beta^\alpha}{(\alpha-k)(\beta+x)^{\alpha-k}}\right]_{x=0}^\infty = 0 - \frac{-\alpha\beta^\alpha}{(\alpha-k)\beta^{\alpha-k}} = \frac{\alpha\beta^k}{\alpha-k}.$$ However, it is worthwhile to observe that $$\frac{\alpha \beta^\alpha}{(\beta+x)^{\alpha+1-k}} = \frac{\alpha \beta^k}{\alpha-k} \cdot \frac{(\alpha-k) \beta^{\alpha-k}}{(\beta+x)^{(\alpha-k)+1}} = \frac{\alpha \beta^k}{\alpha-k} f_{X^*}(x),$$ where $X^*$ belongs to the same parametric family as $X$, except with parameter $\alpha^* = \alpha-k$. In other words, $f_{X^*}(x)$ is a probability density, and it integrates to $1$ over its support. Thus $$\operatorname{E}[(X+\beta)^k] = \frac{\alpha \beta^k}{\alpha-k},$$ whenever $k < \alpha$, as we found with the traditional approach. Consequently, $$\operatorname{E}[X] = \operatorname{E}[X+\beta] - \beta = \frac{\alpha\beta}{\alpha-1} - \beta = \frac{\beta}{\alpha-1},$$ and $$\begin{align*} \operatorname{Var}[X] &= \operatorname{E}[X^2] - \operatorname{E}[X]^2 \\ &= \operatorname{E}[(X+\beta)^2 - 2\beta X + \beta^2] - \operatorname{E}[X]^2 \\ &= \operatorname{E}[(X+\beta)^2] - 2\beta \operatorname{E}[X] - \beta^2 - \operatorname{E}[X]^2 \\ &= \frac{\alpha \beta^2}{\alpha-2} - \frac{2\beta^2}{\alpha-1} - \beta^2 - \left(\frac{\beta}{\alpha-1}\right)^2 \\ &= \frac{\alpha\beta^2}{(\alpha-1)^2 (\alpha-2)}. \end{align*}$$

In regard to the method of moments, the idea is to equate the sample moments with the population moments, until a unique solution for the parameters is determined. Specifically, we would write $$\operatorname{E}[X^k] = \frac{1}{n} \sum_{i=1}^n X_i^k = \overline{X^k}$$ where $(X_1, \ldots, X_n)$ is the sample, and we get a series of equations for each $k = 1, 2, \ldots$. In our case, there are two parameters $\alpha$, $\beta$, so we expect to need to equate only the first two moments ($k \in \{1, 2\}$). This gives us $$\begin{align*} \frac{\beta}{\alpha-1} &= \bar X, \\ \frac{2 \beta^2}{(\alpha-1)(\alpha-2)} &= \overline{X^2}. \end{align*}$$ (It is not the variance we used, but the second moment $\operatorname{E}[X^2]$, for which I did not show the calculation, as it is embedded in the variance calculation above.) This system in $\alpha$, $\beta$ can be solved by solving the first equation for $\beta$, then substituting into the second and solving for $\alpha$: $$\overline{X^2} = \frac{2(\alpha-1)^2 (\bar X)^2}{(\alpha-1)(\alpha-2)} = 2 (\bar X)^2 \frac{\alpha-1}{\alpha-2}.$$ It follows that $$\tilde\alpha_{MM} = \frac{2(\overline{X^2} - (\bar X)^2)}{\overline{X^2} - 2(\bar X)^2}, \quad \tilde\beta_{MM} = \frac{\overline{X^2} \bar X}{\overline{X^2} - 2(\bar X)^2}$$ are the method of moments estimators, based on equating the raw moments.

But this is not the only method! You can also equate on the central moments; e.g., solve the system $$\frac{\beta}{\alpha-1} = \operatorname{E}[X] = \frac{1}{n} \sum_{i=1}^n X_i = \hat \mu, \\ \frac{\alpha \beta^2}{(\alpha-1)^2 (\alpha-2)} = \operatorname{Var}[X] = \frac{1}{n} \sum_{i=1}^n (X_i - \bar X)^2 = \hat \sigma^2.$$ If you do this, you find yet another method of moments estimators: $$\hat \alpha_{MM} = \frac{2\hat\sigma^2}{\hat \sigma^2 - \hat \mu^2}, \quad \hat \beta_{MM} = \frac{\hat \mu(\hat \mu^2 + \hat \sigma^2)}{\hat \sigma^2 - \hat \mu^2}.$$

The question I leave as an exercise for the reader is this: are these actually the same; i.e., is $\tilde \alpha_{MM} = \hat \alpha_{MM}$ and $\tilde \beta_{MM} = \hat \beta_{MM}$? If they are different, how does each perform with respect to the usual properties of estimators such as bias, variance, and mean squared error?

$\endgroup$
  • $\begingroup$ So using that I'll get $\int (u-\beta)u^{-\alpha-1}du = \int_{\beta}^{\infty} u^{-\alpha}du -\beta \int_{\beta}^{\infty} u^{-\alpha-1}du = \frac{u^{-\alpha+1}}{1-\alpha} \mid_{\beta}^{\infty} + \beta \frac{u^{-\alpha}}{\alpha} \mid_{\beta}^{\infty}.$ Can/should I comput the limit (infinity) with L'Hopital? $\endgroup$ – ducks17 Mar 29 '17 at 8:18
  • 1
    $\begingroup$ It is not in indeterminate form and you do not need to apply the L'Hopital rule. $\endgroup$ – BGM Apr 5 '17 at 5:31
  • $\begingroup$ Is what you said "However it is worthwhile to observe that..." the reason why one must assume $\alpha\gt 2$ when finding the method of moments estimators? And did you find the non-central moment since it makes the integration more simple? $\endgroup$ – Remy Oct 12 '18 at 9:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.