Suppose $X\sim N(\mu,\sigma)$. I want to find the following probability $P[\mu \ge \theta |x= \theta -c]$ for $c>0$.
In another word, I saw a sample of Normal distribution, $x$, and know that it is smaller than $\theta$. Now I want to know what is the probability that the mean of distribution is larger than $\theta$.
Attempt 1 I suppose, $\mu$ is a random variable and now with the observation $x=\theta-c$, I want to find $P[\mu\ge \theta]$
$$P[\mu \ge \theta |x= \theta -c]= \frac{P[x=\theta-c|\mu\ge\theta]\times P[\mu\ge\theta]}{P[x=\theta -c]}$$ $$=\frac{\int_\theta^\infty P[x=\theta-c|\mu=t] \times P[\mu=t] dt}{P[x=\theta -c]}$$
Now the only thing I know is that $P[x\le \theta -c|\mu=t]=\Theta (\theta -c,N(t,1)$. I don't have prior information about $P[\mu\ge\theta]$ and say I can assume any distribution that makes the analysis easy.
Attempt 2 I don't have any information about $P[\mu\ge \theta]$ or $P[x=\theta-c]$, so is it acceptable to assume they are uniform on their space?
If so I can do the following
$$P[\mu \ge \theta |x= \theta -c]=P[x=\theta-c|\mu\ge\theta]\\ =\int_\theta^\infty \phi(\mu=t,\sigma,\theta -c) dt =\int_\theta^\infty\frac{1}{\sigma\sqrt{2\pi}}e^{\frac{(t-\theta+c)^2}{2\sigma}}\\ =\int_\theta^\infty\frac{1}{\sigma\sqrt{2\pi}}e^{\frac{(-t+\theta-c)^2}{2\sigma}} =\int_\theta^\infty \phi(\mu=\theta,\sigma,t+c) dt\\ =\Theta(\mu=\theta,\sigma,\theta+c)$$
which is just CDF of $N(\theta,\sigma)$ for $\theta-c$.