I have a question that I come across for practicing. Basically the question is this:

Consider a random sample from the normal distribution with unknown mean and variance $Y_i \sim^{i.i.d.} N(\mu, \sigma^2)$ , $i = 1,...,n$ with the joint prior

$p(\mu, \sigma^2) \propto \sigma^{-1} (\sigma^2)^{-(v_0/2 + 1)} exp\{-\frac{1}{2\sigma^2}[v_0 \sigma_0^2 + k_0(\mu_0 - \mu)^2]\}$ and where $\mu_0, k_0, v_0$ and $\sigma_0^2$ are all known.

Derive the exact form of the marginal posterior distributions of $p(\sigma^2 | Y_1,...,Y_n)$ and $p(\mu | Y_1,...,Y_n)$

So I basically try to write the joint posterior as the product of the prior $p(\mu, \sigma^2)$ and the likelihood function which is $\frac{1}{\sqrt{2\pi}\sigma^n} \exp(-\frac{1}{2\sigma^2} \sum(y_i-\mu)^2)$.

So I have this:

$\sigma^{-1} (\sigma^2)^{-(v_0/2 + 1)} exp\{-\frac{1}{2\sigma^2}[v_0 \sigma_0^2] + k_0(\mu_0 - \mu)^2]\} \cdot \frac{1}{\sqrt{2\pi}\sigma^n} \exp(-\frac{1}{2\sigma^2} \sum(y_i-\mu)^2)$

and then to find $p(\mu | Y_1,...,Y_n)$, I tried taking the integral of the above function from negative infinity to positive infinity.

However I got stuck.

Also I just want to know if it is correct that when I try to find the marginal of $\mu$, i.e. $p(\mu | Y_1,...,Y_n)$: I would take the integral from 0

And when I try to find the marginal of $\sigma^2$, i.e. $p(\sigma^2 | Y_1,...,Y_n)$, I would integrate from negative infinity to positive infinity?

Could someone give some suggestions.


  • $\begingroup$ You are missing a power $n$ on the $\sigma$ in front. Otherwise the derivation of both marginals is quite straightforward. $\endgroup$
    – Xi'an
    Nov 9, 2019 at 0:32
  • $\begingroup$ Hello Xian, do you mean the n that I have put in there now? $\endgroup$
    – john_w
    Nov 9, 2019 at 23:42
  • $\begingroup$ Hello, I still couldn't figure out how to find the marginal of $\mu$ yet. I guess I should integrate with respect to $\sigma^2$ from negative infinity to positive infinity? But it seems that it is not trivial. $\endgroup$
    – john_w
    Nov 10, 2019 at 0:29
  • $\begingroup$ Hi john_w, variance cannot be negative, so the integration bounds are supposed to be $0$ and $\infty$. $\endgroup$
    – Konstantin
    Nov 11, 2019 at 15:55
  • $\begingroup$ The thread at stats.stackexchange.com/questions/52906/… appears relevant. $\endgroup$
    – whuber
    Nov 18, 2019 at 15:52

2 Answers 2



Posterior of $\sigma^2|Y_1,..., Y_n$ is an instance of inverse gamma distribution with the probability density

$$ p(\sigma^2|Y_1,...,Y_n) = \frac{\beta^\alpha}{\Gamma(\alpha)} (\sigma^2)^{-\alpha+1}\exp(-\frac{\beta}{\sigma^2}), $$


\begin{align} \alpha:=\frac{\nu_0+n}{2}, \quad& \beta:=\frac{\nu_0\sigma_0^2+n(\hat \sigma^2 + \frac{k_0}{k_0+n}(\bar Y - \mu_0)^2)}{2},\\ \bar Y := \frac{1}{n}\sum_i Y_i, \quad &\hat \sigma^2 := \frac{1}{n}\sum_i(Y_i-\bar Y)^2. \end{align}

The posterior of $\mu|Y_1,...,Y_n$ is a shifted and scaled Student's t-distribution. It's probability density function can be written down as

$$ p(\mu|Y_1,...,Y_n) = \frac{\Gamma \left(\frac{\nu+1}{2} \right)} {\sqrt{\nu\pi\gamma}\,\Gamma \left(\frac{\nu}{2} \right)} \left(1+\frac{(\mu-\mu')^2}{\nu \gamma} \right)^{-\frac{\nu+1}{2}},$$

where $\Gamma(x)$ is the Gamma function, $\nu:=\nu_0+n$ is the number of degrees of freedom, $\gamma:=\frac{\beta}{\alpha(n+k_0)}$ is the scale parameter $\mu' := \frac{n}{n+k_0}\bar Y + \frac{k_0}{n+k_0}\mu_0 $ is the mean of the interim posterior $\mu|\tau,\mathbf{Y}$:

\begin{equation} \mu|\tau,\mathbf{Y} \sim N(\mu', ((n+k_0)\tau)^{-1}). \qquad\qquad (\star) \end{equation}

Detailed steps:

(There is a great variety of sources online, one of which I am reproducing almost without changes: lectures of Michael I. Jordan from Berkeley.)

Derivation is quite straightforward, once we introduce notation $\tau := \frac{1}{\sigma^2}$, $\alpha_0 := \frac{\nu_0}{2}$, $\beta_0 := \alpha_0 \sigma^2_0$, and recognize in the stated prior the following hierarchical model:

\begin{align} Y_i &\sim N(\mu, \tau^{-1})\\ \mu &\sim N(\mu_0, (k_0\tau)^{-1})\\ \tau &\sim Gamma(\alpha_0,\beta_0)\\ \end{align}

where $Gamma$ stands for Gamma distribution with the probability density function

$$p(\tau|\alpha_0,\beta_0) = \frac{\beta_0^{\alpha_0}}{\Gamma(\alpha_0)} \tau^{\alpha_0-1}\exp(-\tau \beta_0).$$

We are looking for the posterior $\mu| \mathbf{Y}$ and $\tau | \mathbf{Y}$, where $\mathbf{Y}:= (Y_1,...,Y_n) $.

Obtaining posterior $\mu|\mathbf{Y}$ is a matter of taking an expectation of pdf of the interim posterior $(\star)$ with respect to the posterior $\tau|\mathbf{Y}$:

\begin{equation} p(\mu|\mathbf{Y}) = \int_0^\infty p(\mu|\tau, \mathbf{Y}) p(\tau|\mathbf{Y})d \tau \quad (*) \end{equation}

So the first step would be to obtain the posterior $\tau| \mathbf{Y}$, which is just the marginal of the joint posterior $\mu,\tau |\mathbf{Y}$:

\begin{align} p(\tau,\mu|\mathbf{Y}) \propto & \prod_i p(Y_i|\mu,\tau)\cdot p(\mu|\tau)\cdot p(\tau) \\ \propto & \tau^{n/2} \exp\left(-\frac{\tau}{2}\sum_i(Y_i-\mu + \bar Y - \bar Y)^2\right) \cdot \tau^{1/2} \exp\left( -\frac{k_0 \tau}{2}(\mu-\mu_0)^2 \right) \cdot \tau^{\alpha_0-1} \exp(-\beta_0\tau) \\ \propto & \tau^{\alpha_0 + \frac{n}{2}-1}\exp\left(-\tau(\beta_0 + \frac{1}{2}\sum(Y_i-\bar Y)^2) \right) \tau^{1/2} \cdot \exp\left(-\frac{\tau}{2}(k_0(\mu-\mu_0)^2+n(\bar Y - \mu)^2)\right) \end{align}

In the last expression we can factorize a kernel of a normal out of the second term (on the right of $\cdot$):

\begin{align} &\exp\left(-\frac{\tau}{2}(k_0(\mu-\mu_0)^2+n(\bar Y - \mu)^2)\right) =\\ &= \exp\left(-\frac{\tau}{2}((k_0+n)\mu^2-2(k_0\mu_0 + n\bar Y) \mu + k_0 \mu_0^2+n\bar Y^2)\right)\\ &= \exp\left(-\frac{\tau}{2}((k_0+n)(\mu^2-2\frac{k_0\mu_0 + n\bar Y}{k_0+n}\mu + {\mu'}^2) - (k_0+n){\mu'}^2 + k_0 \mu_0^2+n\bar Y^2)\right)\\ &= \exp(-\frac{\tau}{2}(k_0+n)(\mu - \mu')^2) \cdot \exp\left(\frac{\tau}{2}( \frac{k_0^2 \mu_0^2 + 2 n k_0 \mu_0 \bar Y + n^2 \bar Y^2}{n+k_0} - k_0 \mu_0^2 -n \bar Y^2)\right)\\ &= \tau^{1/2} \exp(-\frac{\tau}{2}(k_0+n)(\mu - \mu')^2) \cdot \tau^{-1/2}\exp\left(-\frac{n k_0 \tau}{2(n+k_0)} (\bar Y -\mu_0)^2\right) \end{align}

The first term in the above product is going to integrate to $\sqrt{\frac{2\pi}{k_0+n}}$ (pdf of $N(\mu', \frac{1}{(n+k_0)\tau})$) and may be neglected, whereas the second term will be factorized leaving us with the following posterior for $\tau| \mathbf{Y}$:

\begin{equation} p(\tau|\mathbf{Y}) \propto \tau^{\alpha_0 + \frac{n}{2}-1} \exp(-\tau \left(\beta_0 + \frac{1}{2}\sum_i(Y_i-\bar Y)^2 + \frac{n k_0}{2(n+k_0)}(\bar Y -\mu_0)^2\right)) \end{equation}

in which the kernel of a Gamma distribution is easily recognizable, i.e.

$$\tau|\mathbf{Y} \sim Gamma(\alpha,\beta)$$

where $\alpha := \alpha_0 + \frac{n}{2}$ and $\beta := \beta_0 + \frac{1}{2}\sum_i(Y_i-\bar Y)^2 + \frac{n k_0}{2(n+k_0)}(\bar Y -\mu_0)^2$.

Finally compute the expectation $(*)$:

\begin{align} p(\mu|\mathbf{Y}) = & \int_0^\infty \frac{{\beta}^{\alpha}}{\Gamma(\alpha)}\tau^{\alpha-1} \exp(-\tau \beta) \cdot \frac{(n+k_0)^{1/2}\tau^{1/2}}{\sqrt{2 \pi}} \exp\left(-\frac{n+k_0}{2}\tau (\mu - \mu')^2 \right) d\tau \\ \propto & \int_0^\infty \tau^{\alpha+\frac{1}{2}-1} \exp\left(-\tau \beta - \tau \frac{n+k_0}{2}(\mu - \mu')^2 \right) d\tau \quad (**)\\ \propto & \Gamma(\alpha + \frac{1}{2}) \left(\beta + \frac{n+k_0}{2}(\mu-\mu')^2\right)^{-\alpha-\frac{1}{2}} \\ \propto & (1 + \frac{1}{2\alpha}\frac{(\mu-\mu')^2}{\frac{\beta}{(n+k_0)\alpha}})^{-\frac{2\alpha+1}{2}} \end{align}

In the integrand in expression $(**)$ we see the kernel of $Gamma(\alpha+\frac{1}{2}, \beta + \frac{n+k_0}{2}(\mu - \mu')^2)$ which integrates to the expression in which one can easily recognize the kernel of a Student's t-distribution with mean $\mu'$, scale parameter $\frac{\beta}{(n+k_0)\alpha}$ and $2\alpha$ degrees of freedom.

  • $\begingroup$ hello, could you explain a little more about how you get the term right after you said "being a kernel of a normal integrates (up to a constant factor) to". Sorry I mean how does the 2nd term in the product (i.e the term after the * ) becomes this term (the term I am referring to here) when you integrate with respect to $\mu$? I think I might have to group some terms to make that a Normal Distributed density function with certain means and variances, but I have not figured out how to group the terms yet. $\endgroup$
    – john_w
    Nov 15, 2019 at 21:49
  • $\begingroup$ Hey, sure. I added a clarification, 5 additional lines in the middle. To get to the fourth from the third, you have to just carefully open the parentheses. $\endgroup$
    – Konstantin
    Nov 15, 2019 at 22:54
  • $\begingroup$ Hi, sorry may I know what is the $\bar{\mu}$? $\endgroup$
    – john_w
    Nov 16, 2019 at 0:38
  • $\begingroup$ Also sorry how does the $2n\bar{Y}\mu$ in the second line disappears in the third line in your newly additionally added 5 lines? $\endgroup$
    – john_w
    Nov 16, 2019 at 0:52
  • $\begingroup$ $\bar \mu$ was a mistake aon my behalf, it is really $\mu' = \frac{n\bar Y + k_0 \mu_0}{n+k_0}$. I also added one more intermediate line, hope it makes everything clearer $\endgroup$
    – Konstantin
    Nov 16, 2019 at 11:01

For this type of analysis, it is often possible to decompose the posterior density into a part representing the marginal posterior of one of the parameters, and another part representing the conditional posterior of the other parameter. It turns out to be possible to do this in the present case.

To facilitate our analysis, let us define the useful posterior quantities:

$$\mu_* \equiv \frac{n \bar{y} + k_0 \mu_0}{n + k_0} \quad \quad \quad \quad \quad \beta_* \equiv \frac{||\mathbf{y}||^2 + v_0 \sigma_0^2 + k_0 \mu_0^2 - (n+k_0) \mu_*}{2}.$$

Now, we can solve this problem by writing out the posterior kernel, and then collect all terms involving $\mu$ and simplify this into the kernel of a known density function (in this case the normal density). Using the method of completing the square, we obtain:

$$\begin{equation} \begin{aligned} p(\mu, \sigma^2 | \mathbf{y}) &\propto L_\mathbf{y}(\mu, \sigma^2) \cdot p(\mu, \sigma^2) \\[6pt] &\propto \sigma^{-n} \exp \Bigg( - \frac{1}{2 \sigma^2} \cdot \sum_{i=1}^n ( y_i-\mu)^2 \Bigg) \cdot \sigma^{-v_0 -3} \exp \Bigg( -\frac{1}{2\sigma^2} [v_0 \sigma_0^2+ k_0(\mu_0 - \mu)^2] \Bigg) \\[6pt] &= \sigma^{-n-v_0 -3} \exp \Bigg( - \frac{1}{2 \sigma^2} \Bigg[ \sum_{i=1}^n ( y_i-\mu)^2 +v_0 \sigma_0^2+ k_0(\mu_0 - \mu)^2 \Bigg] \Bigg) \\[6pt] &= \sigma^{-n-v_0 -3} \exp \Bigg( - \frac{1}{2 \sigma^2} \Bigg[ (||\mathbf{y}||^2 -2 n \bar{y} \mu + n \mu^2) + v_0 \sigma_0^2 + (k_0 \mu_0^2 - 2 k_0 \mu_0 \mu + k_0 \mu^2) \Bigg] \Bigg) \\[6pt] &= \sigma^{-n-v_0 -3} \exp \Bigg( - \frac{1}{2 \sigma^2} \Bigg[ -2 (n \bar{y} + k_0 \mu_0 ) \mu + (n + k_0) \mu^2 + ||\mathbf{y}||^2 + v_0 \sigma_0^2 + k_0 \mu_0^2 \Bigg] \Bigg) \\[6pt] &= \sigma^{-n-v_0 -3} \exp \Bigg( - \frac{n + k_0}{2 \sigma^2} \Bigg[ -2 \mu_* \mu + \mu^2 \Bigg] - \frac{||\mathbf{y}||^2 + v_0 \sigma_0^2 + k_0 \mu_0^2}{2 \sigma^2} \Bigg) \\[6pt] &= \sigma^{-n-v_0 -3} \exp \Bigg( - \frac{n + k_0}{2 \sigma^2} \Bigg[ \mu_*^2 -2 \mu_* \mu + \mu^2 \Bigg] - \frac{\beta_*}{\sigma^2} \Bigg) \\[6pt] &= \sigma^{-n-v_0 -3} \exp \Bigg( - \frac{n + k_0}{2 \sigma^2} ( \mu - \mu_* )^2 - \frac{\beta_*}{\sigma^2} \Bigg) \\[6pt] &= \sigma^{-1} \exp \Bigg( - \frac{n + k_0}{2 \sigma^2} ( \mu - \mu_* )^2 \Bigg) \cdot \sigma^{-n-v_0 -2} \exp \Bigg( - \frac{\beta_*}{\sigma^2} \Bigg) \\[6pt] &= \sigma^{-1} \exp \Bigg( - \frac{n + k_0}{2 \sigma^2} ( \mu - \mu_* )^2 \Bigg) \cdot (\sigma^2)^{-(n+v_0)/2 -1} \exp \Bigg( - \frac{\beta_*}{\sigma^2} \Bigg) \\[6pt] &\propto \text{N} \Big( \mu \Big| \mu_*, \frac{\sigma^2}{n + k_0} \Big) \cdot \text{InvGa} \Big( \sigma^2 \Big| \frac{n+v_0}{2}, \beta_* \Big). \\[6pt] \end{aligned} \end{equation}$$

Since the joint density is a probability density, we then have:

$$p(\mu, \sigma^2 | \mathbf{y}) = \text{N} \Big( \mu \Big| \mu_*, \frac{\sigma^2}{n + k_0} \Big) \cdot \text{InvGa} \Big( \sigma^2 \Big| \frac{n+v_0}{2}, \beta_* \Big). $$

From this equation we obtain the marginal distribution:

$$\sigma^2 | \mathbf{y} \sim \text{InvGa} \Big( \frac{n+v_0}{2}, \beta_* \Big). $$

  • $\begingroup$ hello Reinstate Monica, I actually have a question about your answer in almost the second last line of your detailed proof, may I know how you get the $\sigma ^ {-n -v_0 -2} exp(\frac{-\beta_*}{\sigma^2})$ to have the inverse gamma with the first parameter being $n+v_0 + 1$? $\endgroup$
    – john_w
    Dec 14, 2019 at 22:13
  • $\begingroup$ because there must be a term of $\sigma^2$ to the left of the $exp(\frac{-\beta_*}{\sigma^2})$ $\endgroup$
    – john_w
    Dec 14, 2019 at 22:14
  • $\begingroup$ @john_w: Well spotted --- edited to fix error. $\endgroup$
    – Ben
    Dec 14, 2019 at 22:50
  • $\begingroup$ now that when I go over all answers more slowly, I think actually your answer is more close to what I need because it seems that you are going from unknown to known (to actually answer the question from what given in question). But sorry I already awarded the points, should have split if possible or award to your answer. But my other question is your answer and the answer that I awarded the point both got InverseGamma for the posterior of sigma, but the answer for the posterior of mu is different. The other answer has student-t as the marginal of the posterior of mu. $\endgroup$
    – john_w
    Dec 14, 2019 at 23:12
  • $\begingroup$ I now think your answer seems more likely to be correct. May I know do you think the marginal posterior is normal instead of student-t? because it seems it is normal distributed based on your detailed and seem to be correct steps. $\endgroup$
    – john_w
    Dec 14, 2019 at 23:13

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