# Distribution of pivotal quantity

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?

• Sorry, did not initially specify support. Added above. – PatternMatching Apr 14 '15 at 11:55
• 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 $F$? – Glen_b Apr 14 '15 at 12:12
• Again, I was unclear. I mean to construct a CI for $\theta$ given that observations are iid from distribution specified. – PatternMatching Apr 14 '15 at 12:14
• Note that $\bar{x}/\theta$ isn't the only pivot one might construct; all manner of functions of ($x_i/\theta$) might be considered. – Glen_b Apr 14 '15 at 12:17
• 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. – 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 $\text {Beta} (2,1)$.