Consider the posterior distribution derived in the worked example from the previous lecture.
To recap, our parameter of interest is \lambda, prior distribution \textsf{Exp}(a), and likelihood \textsf{Poiss}(\lambda ) for n observations X_1, \ldots, X_ n. This is a Gamma distribution with parameters q_0 and \lambda _0 that you must get from the last two answerboxes in Worked Example Part II.
As before, recall the Gamma distribution , which is a probability distribution with parameters \, q>0 \, and \, \lambda >0 \,, has support on \, (0,\infty ) \,, and whose density is given by \, \displaystyle f(x)=\frac{\lambda ^ q x^{q-1}e^{-\lambda x}}{\Gamma (q)} \,. Here, \, \Gamma \, is the Euler Gamma function.
Of the four sample statistics (mean, median, mode, variance), the Gamma distribution has a simple closed form for three of them. Look up statistics for the Gamma distribution, then for the three that have a simple closed form, calculate them and express your answer in terms of a, n, and \displaystyle \sum _{i=1}^{n} X_ i (use SumXi), otherwise enter -1.
Note that depending on your source, the format of the Gamma distribution may be different, so you must make sure that you have the correct corresponding parameters.
Find the:
Mean?
Median?
Mode?
Variance?
The mean of a Gamma distribution with parameters \(\lambda\) and \(q\) is given by \(\frac{q}{\lambda}\). Therefore, the mean in this case is \(E[\lambda] = \frac{q_0}{\lambda_0}\).
The median of a Gamma distribution is generally not expressed in a simple closed form. Therefore, the median cannot be calculated directly and should be entered as -1.
The mode of a Gamma distribution with parameters \(\lambda\) and \(q\) is given by \(\frac{q-1}{\lambda}\). Therefore, the mode in this case is \(\frac{q_0-1}{\lambda_0}\).
The variance of a Gamma distribution with parameters \(\lambda\) and \(q\) is given by \(\frac{q}{\lambda^2}\). Therefore, the variance in this case is \(Var[\lambda] = \frac{q_0}{\lambda_0^2}\).
In summary:
Mean: \(\frac{q_0}{\lambda_0}\)
Median: -1
Mode: \(\frac{q_0-1}{\lambda_0}\)
Variance: \(\frac{q_0}{\lambda_0^2}\)
The mean of a Gamma distribution with parameters q and λ is given by q/λ.
The median of a Gamma distribution with parameters q and λ can be found by solving the equation for x in the cumulative distribution function (CDF) of the Gamma distribution. Let F(x) be the CDF of the Gamma distribution, then the median is the value of x such that F(x) = 0.5. Unfortunately, there is no closed-form solution for this equation, so we cannot express the median in terms of a, n, and ∑Xᵢ.
The mode of a Gamma distribution with parameters q and λ is (q-1)/λ.
The variance of a Gamma distribution with parameters q and λ is given by q/λ².
Therefore, the mean is q₀/λ₀, the median cannot be expressed in terms of a, n, and ∑Xi, the mode is (q₀-1)/λ₀, and the variance is q₀/λ₀².