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 $x \geq 0$.


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

How can I proceed now?


1 Answer 1


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?

  • $\begingroup$ 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, 2017 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, 2017 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, 2018 at 9:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .