Using posterior expected loss to make a decision between two pdfs

Question

I am trying to solve a question from a past exam paper.

Suppose you have a single observation $X$ from a continuous distribution for which the probability density function (pdf) is either $f_0$ or $f_1$, where

$$ f_0(x) = \begin{cases} 1, & \text{for 0} < x < 1 \\ 0, & \text{otherwise} \end{cases} $$

$$ f_1(x) = \begin{cases} 4x^3, & \text{for 0} < x < 1 \\ 0, & \text{otherwise} \end{cases} $$

On the basis of one observation, we would like to determine whether $f_0$ or $f_1$ is the correct pdf. Suppose that the prior probability that $f_0$ is correct is $\frac{2}{3}$ and that the prior probability that $f_1$ is correct is $\frac{1}{3}$. Suppose that the loss we incur for a correct decision is 0. Assume that the loss from deciding $f_1$ is correct when $f_0$ is in fact true is 1 unit, and that the loss from deciding $f_0$ is correct when $f_1$ is in fact true is 4 units.

If the posterior expected loss needs to be minimised, for what value of $X$ should you conclude that $f_0$ is correct?

I know that the prior probabilities are:

$\pi({f_0}) = \frac{2}{3}$ and $\pi({f_1}) = \frac{1}{3}$

There is only one observation so I think the probabilities are:

$p_{f_0}(x) = 1 $ and $p_{f_1}(x) = 4x^3$

So I think the posterior probabilities are:

$\pi({f_0}|x) \propto \frac{2}{3} \times 1 = \frac{2}{3} $ and $\pi({f_1}|x) \propto 4x^3 \times \frac{1}{3} = \frac{4}{3}x^3$

I can set up a decision function:

Let $d_0$ be the decision that $f_0$ is correct and $d_1$ be the decision $f_1$ is correct. Then $Loss(d_0) = 4$ if $d_0$ is wrong and $Loss(d_1) = 1$ if $d_1$ is wrong.

At this point I am stuck. How can I calculate the posterior expected losses for each of these decisions and then find the value of X that allows me to conclude $f_0$ is correct?

Note: I originally posted on this on CrossValidated but I've deleted that one and re-posted here instead.

Answer 1

We can choose $f_0$ if:

$\pi(f_0|x) \times d_0 < \pi(f_1|x) \times d_1 $

As you stated, $\pi(f_0|x) \propto \pi(x|f_0) \times \pi(f_0) $.

Hence, we can have the following inequality :

$\pi(x|f_0) \times \pi(f_0) \times d_0 < \pi(x|f_1) \times \pi(f_1) \times d_1$

$ 1 \times 2/3 \times 4 < 4 x^3 \times 1/3 \times 1 $

Thus, $ 2 < x^3 $

Since the cube root function is increasing, $ x > \root3\of2\ $