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Suppose that we are to obtain a single observation Y from an exponential distribution with mean theta.
a) Find the distribution of U = Y/theta? Is U a pivotal quantity?
b) Use U = Y/theta and the pivotal method, to form a confidence interval formula for theta with a confidence level of 90%.
I know that the exponential distribution is:
f(x) = (1/theta) * e^(-x/theta)
What I'm confused about is how to find a distribution of U = Y/theta and if it is a pivotal quantity. Any help would be appreciated.
Y is exponential with parameter $\theta$ i.e. $\large p_Y(y)=\frac 1 \theta e^{-y/\theta}$
$Pr[Y<y] = F_Y(y)=1-e^{-y/\theta}$
$U=\dfrac Y \theta$
$Pr[U < u] = Pr[\dfrac Y \theta < u] = Pr[Y< u \theta]$
$Pr[Y< u \theta]=$ $\large 1-e^{-\frac{u \theta}{\theta}}=1 - e^{-u}$
$p_U(u) = \dfrac{d}{du} 1 - e^{-u}= e^{-u}$
Thus $p_U(u)$ is independent of the parameter $\theta$ and is thus a pivotal quantity.
Actually would this be correct?
For random variable X ~ U(0,θ) we need to construct a 90% confidence interval for θ.
Yn would be the largest order statistic from a sample of size n from this distribution and has a pdf:
f(y) = [n*y(n-1)]/θ for 0 ≤ y ≤ θ
Let U = Yn/θ
f(u) = n*u^(n-1) for 0 ≤ u ≤ 1
F(u) = u^n for 0 ≤ u ≤ 1
Then,
P(a < Yn/θ < b) = 0.90
F(a) = 0.05 and F(b) = 0.95
a^n = 0.05 and b^n = 0.95
giving:
a = the nth root of(0.05)
b = the nth root of (0.95)
so,
P(the nth root of(0.05) < Yn/θ < the nth root of (0.95))
giving the confidence interval:
(Yn/the nth root of(0.05), Yn/the nth root of (0.95))
No. That's not how you do it.
For a 90% confidence interval we can assign 5% to the lower tail and 5% to the upper tail.
The interval will be $[a,b]$ such that
$F_U(a)=0.05$
$F_U(b)=0.95$
$F_U(u) = 1-e^{-u}$ as determined earlier. So
$1-e^{-a}=0.05$
$0.95 = e^{-a}$
$a=-\ln(0.95) \approx 0.0513$
$1-e^{-b}=0.95$
$0.05 = e^{-b}$
$b=-\ln(0.05)\approx 2.9957$
So a 90% confidence interval for U is $[0.0513, 2.9957]$
Now $\theta =\dfrac Y U$ so our confidence interval for $\theta$ is given by $\left[ \dfrac Y {2.9957},\dfrac Y {0.0513}\right]$
Note this is not the only $90\%$ confidence interval. Since the distribution isn't symmetric picking one interval over the other is a bit problematic. It's common to make the tails the same probability.