Let a random variable with posterior distribution be given by $X \sim \textrm{Gamma}(10,12)$. This is the result of Jeffreys prior $\pi_J(\lambda)$ multiplied by the Likelihood of i.i.d. $X_1,...,X_n \sim \textrm{Exp}(\lambda)$.

Also, $n=10$ and $\sum_{i=1}^{10} X_i = 12$.

I'm trying to compute a one-sided Bayesian confidence region with level $0.05$ that takes the form $[\alpha,\infty)$, i.e. finding $\alpha$ such that $P(X≥a)=0.95$. This is what I have so far,

$$P(X≥a)=Q(10,\alpha/12) \Rightarrow 12Q^{-1}(10,19/12) \approx 65.1049$$

Where $Q$ is the regularized incomplete gamma function.

The value seems very high to me! Is this actually correct?

This is actually no the part I'm struggling with. I can't manage to compute a one-sided Bayesian confidence region with level $0.05$ that takes the form $(0,b)$, i.e. finding $\alpha$ such that $P(0<X<b)=0.95$. I don't know how to express this as a regularized incomplete gamma function nor inverse cumulative distribution function. If I run some python code, the program eventually fails since it's dealing with both $b$ and $\exp(b)$ which are conflicting and cannot be solved for simultaneously.

Can someone please help me compute $b$?

  • 1
    $\begingroup$ By "Bayesian confidence region" do you instead intend a credible interval? ... en.wikipedia.org/wiki/Credible_interval $\endgroup$
    – Glen_b
    Commented Dec 18, 2022 at 7:11
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    $\begingroup$ Then (in relation to the use of 'confidence') it seems the error is not yours but someone else's. That now raises another question. Is this coursework (or other self-study)? If so, please see the self-study tag wiki info ... stats.stackexchange.com/tags/self-study/info $\endgroup$
    – Glen_b
    Commented Dec 18, 2022 at 7:17
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    $\begingroup$ If I am ready to give an answer, rest assured you will have it. It's not like I am waiting to be prompted in order to begin answering the question. I want to be sure I am clear what is being asked. It would help if you would clarify which parameterization you're using. Can you clarify what $\alpha$ is in $P(X≥a)=Q(10,\alpha/12)$ and where $a$ went? $\endgroup$
    – Glen_b
    Commented Dec 18, 2022 at 7:24
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    $\begingroup$ Please add any info provided in the comments in the question body. Also as Glen_b said, please add the self-study tag. $\endgroup$ Commented Dec 18, 2022 at 8:10
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    $\begingroup$ Done @User1865345 :) $\endgroup$
    – pecer10012
    Commented Dec 18, 2022 at 8:11

1 Answer 1


It appears that you're using the shape-rate parameterization of the gamma. Presumably also $\lambda$ is the rate parameter of the exponential. These will be taken as given for what follows.

Let $\underline{x} = (x_1,x_2,...,x_n)$ be the observed values of $X_1,...,X_n\stackrel{\text{iid}}{\sim}\text{Exp}(\lambda)$. Abusing notation a little, and noting that the Jeffrey's prior for the exponential parameter is $\propto 1/\lambda$ on the positive half line (an improper prior), we have

\begin{eqnarray*} f_{\lambda|\underline{X}=\underline{x}} &\propto& f_{\underline{X}|\lambda}(\underline{x}) \cdot f_\lambda\\ & \propto & \left[\prod_i \lambda e^{-\lambda x_i}\right]\cdot \lambda^{-1} \: I_{(0,\infty)}\\ & = & \lambda^{n-1} e^{-\lambda\sum_i x_i} \: I_{(0,\infty)} \end{eqnarray*}

which we can recognize as the kernel of a $\text{Gamma}(n,\sum_i x_i)$ in the shape-rate parameterization. This much you already knew (though there were multiple issues with the way you were writing it that were causing me some confusion).

Now, a 95% credible interval for $\lambda$ of the form $[a,\infty)$ (again, abusing notation) puts $a$ at the 5th percentile of the distribution. If you have the inverse gamma cdf function, this is simple; with $n=10$ and $\sum_i x_i=12$ you get the value $a\approx 0.4521$ (in R this is qgamma(.05,10,12)). This value makes sense.

(The value $65.1$ you had came from using the wrong parameterization)

With the usual (lower) regularized incomplete gamma, you have to find $a$ such that


Assuming by $Q$ you are looking at the upper rather than the lower regularized incomplete gamma, as with the notation here:


and equivalently the upper regularized gamma here


you'd instead be solving $Q(as,n)=0.95$. That is, finding $a$ such that $Q(12a,10) = 0.95$. This should give the $0.452$-ish value above.

For the other tail, you'd solve $Q(12b,10) = 0.05$

From the values you give, it's not quite clear to me how your inverse function $Q^{-1}$ is set up.

  • $\begingroup$ Thanks for your answer @Glen_b. I'm getting 0.468307637002674 when solving for $b$ in $𝑄(12𝑏,10)=0.05$. This is wrong. Any idea why? $\endgroup$
    – pecer10012
    Commented Dec 18, 2022 at 13:39
  • $\begingroup$ It looks like your Q function's arguments are not in the same order as the definition I used from the links that I gave. Swap them to match the reversed order in your function. You may also need to pay attention to the exact way the Q-function is defined to make sure it is definitely the upper regularized incomplete gamma. $\endgroup$
    – Glen_b
    Commented Dec 18, 2022 at 13:47
  • $\begingroup$ The answer should be roughly 1.3, I think $\endgroup$
    – Glen_b
    Commented Dec 18, 2022 at 13:50
  • $\begingroup$ Yes, that's the correct answer I've got mine accurate to 3dp. The only problem is that the regularized incomplete gamma function returns a univariate polynomial of degree $9$ in terms of $b$ multiplied by $e^{-12b}$ which only wolfram alpha can solve. My python program still breaks as it isn't able to solve such complex equations which, to be honest, really sucks. I'm considering switching to R instead. $\endgroup$
    – pecer10012
    Commented Dec 18, 2022 at 14:02
  • $\begingroup$ wouldn't scipy.stats.gamma.ppf do what you need in python? $\endgroup$
    – Glen_b
    Commented Dec 18, 2022 at 14:28

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