Beta distribution - $\mathbb{P}(\theta \in T|x)$ Consider the following set:
$$T=\left\{\theta \in  \Theta: f(\theta |x) > \sup_{\theta \in \theta_o} f(\theta | x)\right\}.$$
Suppose that $\theta_o =\left\{\frac{3}{4}\right\}$ and $\theta|x \sim \beta (2,5)$. Calculate $\mathbb{P}(\theta \in T | x)$.
I couldn't do much since I didn't understand how can I find this probability.
P.S: $\Theta$ is the parametric space
Thanks in advance!
Edit: My attempt is the following:
$$f(\theta |x) > f(3/4|x) \iff \theta (1-\theta)^4 > \frac{3}{1024}$$
I think we can aproximate the probability of $\theta \in T$ by $\int_{0}^{3/4} 30 \theta(1-\theta)^4d\theta$, since the probability of $\theta \in T$ is the same as the probability of $\theta$ have "more density " than $\frac{3}{4}$. This can be aproximate by the probability of $\theta \in [0, \frac{3}{4}]$

 A: You have identified a correct approach. The only point I am not quite sure of is the logic you used to conclude that θ∈[0,3/4].
I would proceed this way (a bit more analytical):


*

*First, find the max/min whatever you could find from the pdf of the given Beta dstribution(2,5). To do this, I used first order and second order derivatives of θ(1-θ)^4 to determine the function has its max value attained at θ=1/5=0.2. Also, θ(1-θ)^4 being a pdf of a distribution it can never go below 0. 
Observation: the function θ(1-θ)^4 is an increasing function of θ for 0<θ<0.2 and a decreasing function of θ for 0.2<θ<1

*Next, try to gauge the nature of θ(1-θ)^4 around θ=3/4. You will notice θ(1-θ)^4 is a decreasing function for ¾< θ<1 (from the sign of the first order derivative of θ(1-θ)^4 over that range – you’ll see it will be negative)

*Next, think about your goal. You need to find the area under the pdf curve where the pdf takes values more than ¾. Since θ(1-θ)^4 is a decreasing function for ¾< θ<1, the upper bound of the range for the possible values of θ in the set T is ¾

*So, the job remains to find the lower bound of the range of θ in T.

*Also, recall the first step: we know the max of θ(1-θ)^4 is attained at θ=0.2. So, the lower bound will lie between 0 and 0.2. At this point, the problem of finding the lower bound reduces to finding the zero of the polynomial function θ(1-θ)^4 –(3/4^5) between 0 and 0.2. Here, 3/4^5 is nothing but the functional value evaluated at θ=3/4.

*Perhaps apply the concept of bi-section method to find the roots. Just notice that this is equivalent to comparing the signs of (r-1)^4 and 3(r/4)^5 by increasing the r twice each iteration (so that, the root is 1/r). You’ll see at one point the root lies between 1/640 and 1/320 which is pretty small (the root is approx 0.0029)


So, the evaluated probability: 
P(θ∈T|x) = ∫{θ(1-θ)^4} / B(2,5) dθ over T is approx {θ: 0 < θ < 3/4}
[Note: B(2,5) = 1/30]
P(θ∈T|x) = ∫30 θ(1-θ)^4 dθ over T is approx {θ: 0 < θ < 3/4}
That is, P(θ∈T|x) is approx 0.995
Regards,
Sauvik
