1. Pivotal Method

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.

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3. Re: Pivotal Method

How are you being given these questions without being taught the basics behind them?

4. Re: Pivotal Method

Originally Posted by romsek
How are you being given these questions without being taught the basics behind them?
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.

5. Re: Pivotal Method

For part b. would this be the correct answer?

P[ -z(0.05) ≤ (1/θ)*e^(-y/θ) ≤ z(0.05) ] = 0.90

=> P[ -1.64 ≤ (1/θ)*e^(-y/θ) ≤ 1.64 ] = 0.90

=> P[ -1.64 ≤ (1/θ)*e^(-u) ≤ 1.64 ] = 0.90

=> P[ -e^(-u) / 1.64 ≥ θ ≥ e^(u)/1.64 ] = 0.90

??

7. Re: Pivotal Method

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))

8. Re: Pivotal Method

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.