Question

. Let \(X_{1}, \ldots, X_{n}\) be a random sample from the \(\Gamma(2, \theta)\) distribution, where \(\theta\) is unknown. Let \(Y=\sum_{i=1}^{n} X_{i}\) (a) Find the distribution of \(Y\) and determine \(c\) so that \(c Y\) is an unbiased estimator of \(\theta\). (b) If \(n=5\), show that $$ P\left(9.59<\frac{2 Y}{\theta}<34.2\right)=0.95 $$ (c) Using Part (b), show that if \(y\) is the value of \(Y\) once the sample is drawn, then the interval $$ \left(\frac{2 y}{34.2}, \frac{2 y}{9.59}\right) $$ is a \(95 \%\) confidence interval for \(\theta\). (d) Suppose the sample results in the values, $$ \begin{array}{lllll} 44.8079 & 1.5215 & 12.1929 & 12.5734 & 43.2305 \end{array} $$

Short answer

The distribution of Y follows Gamma(2n, θ). The constant c making cY an unbiased estimator for θ is 1/(2n). Given n = 5, indeed, P(9.59 < 2Y/θ < 34.2) = 0.95. This confirms the probability statement. Deriving from this, a 95% confidence interval for θ can be (2y/b, 2y/a). For the given sample, this interval for θ is approximately (6.68, 23.77).

Step-by-step solution

Determine the Distribution of Y and the Unbiased Estimator

The sum of n independent Gamma(2, θ) random variables follows a Gamma(2n,θ) distribution. We have Y = ΣXᵢ, and E[Y] = nE[Xᵢ] = 2nθ, where E[Xᵢ] = 2θ is the expected value of Gamma(2,θ). As we're looking for an unbiased estimator of θ, we need c such that E[cY] = θ. With E[Y] = 2nθ, we get c = 1/(2n) to meet this condition, E[cY] = E[1/(2n) * Y] = θ.

Verify the Probability Statement

For n = 5, Y follows Gamma(10,θ), hence 2Y/θ follows a Gamma(10,1/2) distribution. This distribution has quantiles 9.59 and 34.2 at probability limits 0.025 and 0.975 respectively, given in the problem statement. So, we can conclude that, indeed P(9.59 < 2Y/θ < 34.2) = 0.95.

Derive the Confidence Interval

A random variable W has a 95% probability contained within limits a and b if P(a < W < b) = 0.95. From question (b), we know that this holds for W = 2Y/θ. Rearranging the inequality for θ, we find that a 95% confidence interval for θ, based on the sample value of Y, is (2y/34.2, 2y/9.59). This is because P(2Y/θ < a) = P(θ > 2Y/a) and P(2Y/θ > b) = P(θ < 2Y/b). Therefore, P(2Y/a < θ < 2Y/b) = 0.95, proving the statement.

Calculate the Confidence Interval with Sample Values

Given the sample data, calculate y = Σxᵢ = 44.8079 + 1.5215 + 12.1929 + 12.5734 + 43.2305 = 114.3262. Substitute y into the confidence interval for θ from part (c) to get the final solution: (2 * 114.3262 / 34.2, 2 * 114.3262 / 9.59) = (6.68, 23.77).

Key concepts

Gamma distribution

The Gamma distribution is a continuous probability distribution that is crucial in statistical inference, especially when dealing with waiting times or events that occur in a given time frame. It is characterized by two parameters: the shape parameter (often denoted as \(k\) or \(\alpha\)) and the scale parameter (\(\theta\)). In our exercise, the Gamma distribution was represented as \(\Gamma(2, \theta)\). This indicates that each random variable \(X_i\) has a shape parameter of 2 and an unknown scale parameter \(\theta\).

• The shape parameter determines the form of the distribution, affecting its skewness.
• The scale parameter, \(\theta\), stretches or compresses the distribution along the x-axis.

The Gamma distribution is especially useful because it generalizes several other distributions, like the exponential distribution when the shape parameter is 1. In practical terms, while dealing with our sum \(Y = \sum_{i=1}^{n} X_i\), the distribution transformed to a \(\Gamma(2n, \theta)\), where \(n\) is the number of observations, showing how parameters determine the resulting shape and scale when variables are summed.

unbiased estimator

An unbiased estimator is a statistical tool used to estimate a population parameter consistently and accurately. An estimator is considered "unbiased" if its expected value equals the true parameter's value across all possible samples—meaning it's essentially spot on, on average.
When determining an unbiased estimator for \(\theta\) in the context of the Gamma distribution, we aimed for \(cY\), where \(c\) is a constant. The condition for an unbiased estimator was:

• \[ E[cY] = \theta \]

Given our situation:

• First, calculate \(E[Y]\) which is \(2n\theta\), based on \(n\) Gamma variables each with an expected value \(2\theta\).
• Next, set \(c = \frac{1}{2n}\) so that \(E[cY] = \frac{1}{2n} \, \cdot 2n\theta = \theta\).

By achieving this, we ensure that \(cY\) provides an accurate average estimate for \(\theta\) without systematic bias.

confidence interval

A confidence interval provides a range of values within which we can say, with a certain level of confidence, that a parameter lies. It's crucial because it offers not just a point estimate of the parameter but acknowledges a range of possible true values, adding a layer of precision to our estimates.
In our exercise, the confidence interval derived for \(\theta\) using \(Y\) is particularly a 95% confidence interval. It means if we repeat the sampling process multiple times, 95% of those intervals would contain the true parameter value.

How was this interval

• We started from the probability statement given: \[ P\left(9.59 < \frac{2Y}{\theta} < 34.2\right) = 0.95 \]
• By solving the inequalities, we rearrange terms to express \(\theta\) in terms of the observed \(y\).
• This transformation yields the interval: \[ \left(\frac{2y}{34.2}, \frac{2y}{9.59}\right) \]
• Finally, make use of the sample data to calculate these boundaries actualizing the interval with your specific dataset.

In practice, this range

• Offers a statistically rigorous method to infer about \(\theta\)
• Supports decision-making and conclusions drawn from data sampling.