Suppose we have a point mass prior,
$$\theta \sim \begin{cases} I(\theta=1) ,& prob=\frac{1}{2} \\ Gamma(c,c), & prob=\frac{1}{2} \end{cases}$$
Then if we are asked
$\lim_{c \to \infty} P(\theta=1)$
Now here is the issue, since Gamma is a continuous distribution, it seems that in the case of gamma,we will never have $\theta=1$.
To me it thus seems that regardless of the value of c, the $p(\theta=1)=\frac{1}{2}$
However, we also have that since expected value of a $gamma(a,b)=\frac{a}{b}$ so that the expected value of the gamma is 1 when we have $a=b=c$
But, by Markov, for $X \sim Gamma(c,c)$
$\lim_{c \to \infty} Pr(|X-\mu| \lt \epsilon) \to 1$ for any $\epsilon \gt 0$
So is $\lim_{c \to \infty}P(\theta=1) =1$ , or is $\lim_{c \to \infty}P(\theta=1)=\frac{1}{2}$
As even though the markov inequality holds, it is a continous distirbution, so we will never have it exactly equal to 1.
Thanks all