• Method 1: Using PDF

To show that the pdf of a random variable $\theta$ is proper, we need to show it satisfies the definition of a probability density function by showing

$$\int_{\Theta}f(\theta)\mathrm{d}\theta=1.$$

We can express the density in terms of its kernel as $f(\theta)=C\kappa(\theta)$. Then we need to show

$$\int_{\Theta}C\kappa(\theta)\mathrm{d}\theta=1\implies\int_{\Theta}\kappa(\theta)\mathrm{d}\theta=\frac{1}{C}.$$

We require $C\neq0$ so that the density is non-degenerate. Then $1/C<\infty$ and we only need to show

$$\int_{\Theta}\kappa(\theta)\mathrm{d}\theta<\infty,$$

because $C$ is explicitly determined by $C=1/\int_{\Theta}\kappa(\theta)\mathrm{d}\theta$.

• Method 2: Using representation

If the posterior of a parameter $\theta$ can be expressed in terms of known distributions, then it is proper by construction.

• (Step 1) Derive $[\pi_0, \pi_1 \mid x_{1:m}^{(0)}, x_{1:n}^{(1)}]$.
• (Step 2) Since $\theta$ is a function of $\pi_0$ and $\pi_1$, you may use representation to define the posterior of $\theta$.

• (Step 1) Suppose we can prove that $f(p_1, p_2 \mid x) = \texttt{dnorm}(p_1,x,1)\times\texttt{dexp}(p_2)$.
• (Step 2) Let $\phi = p_1 - p_2$. Then, given $x$, the posterior of $\phi$ can be represented as follows:
  \[
  \phi = A- B,
  \]
  where $A\sim \text{Normal}(x,1)$ and $B\sim\text{Exp}(1)$ are independent.

• (Method 1) Algebraically, you can derive the prior pdf of theta. After some algebras, you can find a simple closed-form solution for that. However, it may take you some energy for deriving it.
• (Method 2) You may use simulation. However, the tail of the prior pdf of theta is very fat. So, you will obtain many crazily large values of theta. In this case, the R-function density(.) is not very stable. There are some sophisticated methods to deal with it. You will learn it in Stat 3005 Nonparametric Statistics. Now, I would like to tell you a simple trick for handling this.
  • Step 1: Simulate $\theta$ as usual.
  • Step 2: Select the $\theta$'s that are smaller than a certainvalue (e.g., 5). Store them as thetaSelected.
  • Step 3: Find the probability that $\theta$ is smaller than 5. Denote it by p.
  • Step 4: Compute the densityof the selected $\theta$'s. In R set, fit=density(thetaSelected).
  • Step 5: Plot the (fit$y)*p against fit$x. It is the correct density value after adjusting for the selection bias.Then you will obtain a graph similar to this:

n = 1e7
x = runif(n)
y = runif(n)
theta = x/y-1
theta0 = theta[which(abs(theta)<5)]
p = mean(abs(theta)<5)
d = density(theta0)
plot(d$x, d$y*p, xlim=c(-1,4), type="l", lwd=2, col="blue", ylab="Density", xlab=expression(theta))
abline(v=c(0-1,1/2-1), h=0, lty=2, col="red")

• Regarding the closed form of the prior:

The prior density is

$$\begin{align}f\left(\theta\right)=\left\{\begin{array}{cc}\frac{1}{2} & -1<\theta<0 \\\frac{1}{2 (\theta+1)^{2}} & \theta\geq0 \end{array}.\right.\end{align}$$

It can be derived by the following steps. Firstly, let $$y_1=\pi_0/\pi_1-1,y_2=\pi_1.$$ Correspondingly, $$\pi_0=(y_1+1)y_2,\pi_1=y_2.$$ Because of the constraint $\pi_0,\pi_1\in(0,1)$, it must be satisfied that $(y_1+1)y_2\in(0,1),y_2\in(0,1)$. Moreover, it is easily seen that $y_1=\pi_0/\pi_1-1\in(-1,\infty)$. Therefore, we must have $$y_1\in(-1,\infty),y_2\in(0,\min(1,1/(y_1+1))).$$

Secondly, the Jacobian matrix of the mapping from $(y_1,y_2)$ to $(\pi_0,\pi_1)$ is
$$J=\left|\begin{array}{cc}y_2 & y_1+1 \\0 & 1 \end{array}\right|=y_2.$$

Thus we have
$$\begin{equation}f(y_1,y_2) = f(\pi_0,\pi_1)\left|J\right| = y_2\mathbb{1}(y_1\in(-1,\infty),y_2\in(0,\min\{1,1/(y_1+1)\})).\end{equation}$$

Thirdly, denoting the support of $y_2$ as $\mathcal{Y}_2$, the marginal distribution of $y_1$ can then be found by
$$\begin{align*} f(y_1) =& \int_{\mathcal{Y}_2}f(y_1,y_2)\mathrm{d}y_2=\left\{\begin{array}{cc}\int_{0}^{1}y_2\mathrm{d}y_2&\text{ if } 1/(y_1+1)>1\\\int_{0}^{1/(y_1+1)}y_2\mathrm{d}y_2 &\text{ if } 1/(y_1+1)\leq1\end{array}\right.\\ =&\left\{\begin{array}{cc} \frac{1}{2}&\text{ if } -1<y_1<0\\\frac{1}{2(y_1+1)^2} &\text{ if } y_1\geq0. \end{array}\right. \end{align*}$$

Since $\theta\equiv y_1$, the desired density follows.

The solution is also updated.

• Regarding the R codes:

Repetition of $0$ is for better visualization. We know $f(\theta)=0 $ on the interval $[-2,-1]$.

And $101$ is the repetition number. The repetition number comes from how we set the x-axis. Notice the interval $(-1,0)$ is divided into 100 sub-intervals with length $0.01$. However, there are $101$ boundary points. Hence there is an extra $1$.

1. You need to mention that $B_0$ and $B_1$ are independent.
2. The meaning of the notation $B_0+B_1$ hasn't been made clear. Usually, when we write $B_0+B_1$, we are referring to a random variable that can be represented as the summation of two Beta random variables. However, $f(\pi_0,\pi_1)$ is a density function with two variates. Can we equate them? Moreover, why is it a summation (of two independent Beta random variables)?