[Math] Find a pivotal quantity and use it to approximate a 95% confidence interval

confidence intervalprobabilityprobability distributionsstatistical-inferencestatistics

Suppose $Y_1, Y_2, … , Y_n$ denotes a random sample from a uniform distribution on the interval $(-\theta, 4\theta)$, where $\theta \gt 0$ us an unknown parameter.

Assume that n is sufficiently large, find a pivotal quantity in terms of $\overline{Y}$and $\theta$. Use this pivotal quantity to derive a formula of:

(i) an approximate 95% confidence interval for $\theta$.

(ii) an approximate 95% confidence interval for $\theta^2$.

What I have tried so far:

(i)
We know that $Y$ ~ $Unif(-\theta, 4\theta)$, so $\mu = 1.5\theta$ and $\sigma^2 = \frac{25\theta^2}{12}$. However, we also know that n is sufficiently large to apply the central limit theorem, so $Y$ ~ $N(1.5\theta, \frac{25\theta^2}{12})$

From that, we can create the pivotal quantity: $$\frac{\overline{Y} – \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{\overline{Y} – 1.5\theta}{\frac{5\theta}{\sqrt{12n}}} $$

So now we can set the problem up as $P(Z_{0.025} \le \frac{(\overline{Y} – 1.5\theta)\sqrt{12n}}{5\theta} \le Z_{0.975})$

After going through the math, I ended up with $P((\frac{-\sqrt{12n}}{9.8} + \frac{2}{3})\overline{Y} \ge \theta \ge (\frac{\sqrt{12n}}{9.8} + \frac{2}{3})\overline{Y})$ as my interval. I'm not really sure if that is correct. It seems a bit messy.

(ii)
For this part, I was having difficulty getting started. I'm not really sure how to go about finding a pivotal quantity for $\theta^2$, or if I'm just supposed to use the above pivotal quantity, which wouldn't make sense to me.

I know that to be a pivotal quantity, it has to be a function of my random variable Y and my parameter $\theta$. And I also know that the distribution of my pivotal quantity has to be parameter free. But I'm otherwise stuck at this point.

Any help or direction would be greatly appreciated.

Best Answer

First note that $Z_{0.025}=-Z_{0.975}=-1.96$ and solve the inequaalities $$-1.96 \le \frac{(\overline{Y} - 1.5\theta)\sqrt{12n}}{5\theta} \le 1.96$$ Multiplying by $5\theta$ and dividing by $\sqrt{12n}$, obtain $$-\frac{1.96\cdot 5\theta}{\sqrt{12n}} \le \overline{Y} - 1.5\theta \le \frac{1.96\cdot 5\theta}{\sqrt{12n}}$$ Add $1.5\theta$ to all parts: $$-\frac{1.96\cdot 5\theta}{\sqrt{12n}}+1.5\theta \le \overline{Y} \le \frac{1.96\cdot 5\theta}{\sqrt{12n}}+1.5\theta$$ Take $\theta$ out of brackets: $$\theta\left(\frac32-\frac{9.8}{\sqrt{12n}}\right) \le \overline{Y} \le \theta\left(\frac32+\frac{9.8}{\sqrt{12n}}\right)$$ Solving both inequalities with respect to $\theta$ for sufficiently large $n$ which guarantee both brackets are positive, we have: $$\frac{\overline{Y}}{\frac32+\frac{9.8}{\sqrt{12n}}}\le\theta\le \frac{\overline{Y}}{\frac32-\frac{9.8}{\sqrt{12n}}} $$ Note that $$\frac{1}{\frac32\pm\frac{9.8}{\sqrt{12n}}}\neq \frac23\pm\frac{\sqrt{12n}}{9.8}$$

To get confidence interval for $\theta^2$, simply square both ends of this interval. The only problem is that $\overline{Y}$ can be negative but for $n$ sufficiently large probabililty of this event is negligible since $\overline{Y}\to 1.5\theta >0$ in probabililty as $n\to\infty$.

Related Solutions

[Math] Find a confidence interval using as pivotal quantity a function of the MLE

Assuming you want a confidence interval for $\theta$, you may use the CLT or a known chi distribution for variances and some integrals to get an interval for the parameter. First, we have that

$$ \hat{\theta}=-\frac{n}{\sum ln\ x_i}. $$

For the expected value we have the integral

$$ E[X] = \int_0^1 \theta x^{(\theta-1)}x\ \text{d}\theta = \frac{\theta}{1+\theta} $$

And for the second moment

$$ \mu_2 = \int_0^1 \theta x^{\theta-1}x^2\ \text{d}\theta = \frac{\theta}{2+\theta} $$

So the variance is

$$ Var(X) = \frac{\theta}{2+\theta} - \left( \frac{\theta}{1+\theta} \right)^2 = \\ = \frac{\theta}{(\theta + 1)^2(\theta+2)}. $$

Now then, we know that the following quotient has a chi-squared distribution, so in this case

$$ C = \frac{(n-1)S^2}{\sigma^2} \sim \chi^2_{n-1} \text{ and } C = \frac{(n-1)S^2}{\frac{\theta}{(\theta + 1)^2(\theta+2)}} $$

where $S^2 = \frac{1}{n-1}\sum(X_i-\overline{X})^2$. Finally we can work with the random variable $C$ distributed as a chi squared. We want a confidence interval of $(1-\alpha)$. Then

$$ P \left( \chi^2_{n-1}(\alpha/2) < C < \chi^2_{n-1}(1-\alpha/2)\right) = 1-\alpha $$

denoting that $\chi^2_{n-1}(\beta)$ is the $\beta$ percentile of the $\chi^2$ which may be found in reference tables.

From here we work with the interval only substituting the percentiles with $\frac{1}{D_l}$ and $\frac{1}{D_s}$, lower and superior.

$$ \frac{1}{D_l} < C < \frac{1}{D_s} \implies \frac{1}{D_l} < \frac{(n-1)S^2}{\frac{\theta}{(\theta + 1)^2(\theta+2)}} < \frac{1}{D_s} \\\ \\\ \\\ D_l > \frac{\frac{\theta}{(\theta + 1)^2(\theta+2)}}{(n-1)S^2} > D_s \\\ \\\ \\\ D_l((n-1)S^2) > \frac{\theta}{(\theta + 1)^2(\theta+2)} > D_s((n-1)S^2) \\\ \\\ \\\ (D_l((n-1)S^2)) > \frac{\theta}{(\theta + 1)^2(\theta+2)} > (D_s((n-1)S^2)) \\\ \\\ \\\ (\hat{\theta} + 1)^2(\hat{\theta}+2)\left( D_l((n-1)S^2) \right) > \theta > (\hat{\theta}+ 1)^2(\hat{\theta}+2)\left( D_s((n-1)S^2) \right) \\\ \\\ \\\ $$

where, taking the before definition and derivation, $\hat{\theta} = -\frac{n}{\sum ln\ x_i}$ and $S^2 = \frac{1}{n-1}\sum(X_i-\overline{X})^2$. When we sent the $\theta$ to the other side, we simply used its estimate to work around the problem. (However a mathematician should confirm this is valid!) Either way, if you find a better known distribution that may include $\theta$, then try it and tell us if it is simpler!

Hope it helps!

Note: made some edits due to an error :/

Confidence Intervals using Pivotal Quantities

For most practical purposes it is not feasible to get a useful CI for $\sigma^2$ from only $n = 6$ observations. My guess is this is a drill problem and you're intended to assume normality of the population and to assume you can get useful information from such a small sample.

Suppose data are normal and both population mean $\mu$ and population variance $\sigma^2$ are unknown, $\mu$ is estimated by $\bar X$ and $\sigma^2$ is estimated by $S^.$ Then a useful pivotal quantity is

$$\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(\nu = n-1).$$

Suppose you want a 90% confidence interval for $\mu$ based on your $n = 6$ observations. Then the values $L = 1.145$ and $U=11.070$ cut 5% of the probability from the lower and upper tails of the $\mathsf{Chisq}(5),$ respectively. These cut-off values can be found from tables of the chi-squared distribution or by using software. In R statistical software, we have the following computation:

qchisq(c(.05,.95), 5)
[1]  1.145476 11.070498

Thus $$0.90 = P\left(L \le \frac{(n-1)S^2}{\sigma^2} \le U\right) = P\left(\frac{(n-1)S^2}{U} \le \sigma^2 \le \frac{(n-1)S^2}{L}\right),$$ so that the desired 90% confidence interval is of the form $$\left( \frac{(n-1)S^2}{U},\; \frac{(n-1)S^2}{L}\right) = (0.0393, 0.3801).$$

For your data, the computation in R amounts to the following:

x = c(9.8, 9.43, 8.97, 9.33, 9.14, 9.55)
df = length(x) - 1
v = var(x)  
[1] 0.08708
df*v/qchisq(c(.95,.05), df)
[1]  0.03932976 0.38010392

Notice that the point estimate $S^2 = 0.0871$ of $\sigma^2$ is included in this confidence interval. However, owing to the skewness of the distribution $\mathsf{Chisq}(5),$ the point estimate is not at the center of the interval. [That is, unlike a CI for $\mu$ based on the symmetrical t distribution and $\bar X,$ this CI for $\sigma^2$ is not of the form 'point estimate plus or minus a margin of error'.]

If you want a 95% CI for the population standard deviation $\sigma,$ then take square roots of both endpoints of the CI for $\sigma^2.$

Best Answer

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[Math] Find a confidence interval using as pivotal quantity a function of the MLE

Confidence Intervals using Pivotal Quantities

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