I'm attempting to determine whether a pivot can be used to construct a confidence interval for $\theta$ given that observations are iid and from the distribution below. Specifically:

$f(x \mid \theta) = \frac{2x}{\theta^2}$ with $0 < x < \theta$ and $\theta > 0$

Since this is a scale family pdf, I would expect the pivot $\frac{\bar{x}}{\theta}$ to be useful. However, it is not apparent how to determine the distribution of the pivot.

Certainly, the distribution of $X$ seems related to the uniform, but I can only seem to find things like the Bates distribution, which gives distribution of the mean of $n$ Uniform(0,1)'s.

Any thoughts?

  • 1
    $\begingroup$ Sorry, did not initially specify support. Added above. $\endgroup$ – PatternMatching Apr 14 '15 at 11:55
  • $\begingroup$ When you say "construct a confidence interval for a given distribution", do you mean something like "construct a confidence interval for $\theta$, given the observations are from the distribution below", or do you mean to actually construct a CI for $F$? $\endgroup$ – Glen_b Apr 14 '15 at 12:12
  • $\begingroup$ Again, I was unclear. I mean to construct a CI for $\theta$ given that observations are iid from distribution specified. $\endgroup$ – PatternMatching Apr 14 '15 at 12:14
  • $\begingroup$ Note that $\bar{x}/\theta$ isn't the only pivot one might construct; all manner of functions of ($x_i/\theta$) might be considered. $\endgroup$ – Glen_b Apr 14 '15 at 12:17
  • $\begingroup$ Agreed. For example, using the transformation $T = X/\theta$ makes things considerably simpler. I guess I'm more curious about the possibility of using the above. $\endgroup$ – PatternMatching Apr 14 '15 at 12:22

Your variable is a scaled Beta- distributed random variable.
Specifically: Consider the random variable $Y$ that follows a $\text {Beta} (\alpha,\beta)$ distribution in $[0,1]$. The general form of the $\text {Beta} (\alpha,\beta)$ density is

$$f_Y(y\mid \alpha, \beta) = \frac {y^{\alpha-1}(1-y)^{\beta-1}}{B(\alpha, \beta)}$$

where $B(\alpha, \beta)$ is the Beta function First we have to get rid of the $(1-y)^{\beta-1}$ term, so we must set $\beta =1$. Doing this we obtain

$$f_Y(y\mid \alpha, 1) = \alpha y^{\alpha-1}$$

Second we want the variable to eventually not be raised to a power,so it follows that we must set $\alpha -1 = 1 \implies \alpha = 2$, and we have arrived at

$$f_Y(y\mid 2, 1) = 2y, \;\;\;y \in [0,1]$$

Consider now the random variable (your random variable)

$$X = \theta Y \implies Y = X/\theta$$

Applying the change-of-variable formula we obtain

$$f_X(x) = \left|\frac {\partial Y}{\partial X}\right|\cdot f_Y(x/\theta) = \frac {1}{\theta} \cdot 2\frac {x}{\theta} = \frac {2x}{\theta^2},\;\;\; x\in (0,\theta)$$

which is the density you have in your hands. Knowing this, we turn to the pivot and obtain

$$T\equiv \frac {\bar X}{\theta} = \frac {1}{\theta}\frac 1n\sum_{i=1}^nX_i = \frac {1}{\theta}\frac 1n\sum_{i=1}^n\theta Y_i = \bar Y$$

So, if you have an i.i.d. sample drawn from the distribution of $X$, the pivot $T$ that you are examining is in reality the sample mean from an i.i.d. sample of $\text {Beta} (2,1)$ RV's, for the true value of $\theta$.

As far as I recall, the sample mean from a Beta-distributed sample does not have a closed-form distribution, so confidence intervals should rely on asymptotics and/or Monte Carlo procedures.

A point estimate is straightforward: since $E(Y) = 2/3$, the method of moments estimator is

$$\hat \theta_{MoM} = \frac 32 \bar X$$

If you have a large sample, you could also obtain the empirical distribution of the $X$'s and then find the value by which you need to divide the observations in order for the empirical frequency distribution to match the theoretical density of a $\text {Beta} (2,1)$.

| cite | improve this answer | |

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.