mjhjmtu

I am stuck in this question:

if $a = 1$ then $m sim U(0.2,1)$ else if $a=0$ then $m sim U(0,0.5)$. The question is if $m$ is $0.3$ what is the probability that $a$ equals to 1?

My thought is to compute $p(a=1mid m=0.3)$ and $p(a=0mid m=0.3)$ and whichever class gives the higher probability then it is the answer. However, when I am executing this thought I have a problem of computing $p(m=0.3mid a=1)$ which is supposed to be zero since it follows $U(0.2,1)$. I feel like I can use the density function to compute this probability but I am not sure why?

If $a=1$ then the density function of $m$ is $displaystyle f_m(x) = begincases 1/(1-0.2) = 1.25 & textif 0.2le xle 1, \ 0 & textif x<0.2 text or x>1. endcases$

If $a=0,$ it is $displaystyle f_m(x) = begincases 1/(0.5-0) = 2 & textif 0le xle0.5, \ 0 & textif x<0 text or x>0.5. endcases$

Thus the likelihood function is
$$
begincases L( 1 mid m=0.3) = 1.25, \ L(0 mid m=0.3) = 2. endcases
$$

Hence the posterior probability distribution is
$$
begincases Pr(a=1mid m=0.3) = ctimes 1.25timesPr(a=1), \[5pt] Pr(a=0mid m = 0.3) = ctimes 2 times Pr(a=0). endcases tag 1
$$
The normalizing constant is
$$
c = frac 1 1.25Pr(a=1) + 2Pr(a=0)
$$
(that is what $c$ must be to make the sum of the two probabilities in line $(1)$ above equal to $1.$)

So for example, if $Pr(a=1)=Pr(a=0) = dfrac 1 2$ then $Pr(a=1mid m=0.3) = dfrac 5 13.$

Instead of Bayes rule, we should look closely at the definition of conditional probability. as Bayes rule is simply a small transformation of the definition of conditional probability.

With discrete probabilities, it is simple to define

$$P(A = a mid B = b) = fracP(A = acap B = b)P(B = b)$$

As you pointed out, this would be ill-defined if both $A$ and $B$ were continuous, as it would result $0/0.$

Instead, let's think about $P(A in a pm varepsilon mid B in b pm varepsilon)$ for a continuous distribution. Then the value

$$ fracP(A in a pm varepsilon cap B in b pm varepsilon)P(B in b pm varepsilon) $$

is properly defined for all $epsilon$, as long as $int_b - varepsilon^b + varepsilon f_b(x) , dx > 0 $. Finally, we just define the conditional distribution of $A|B$ as

$$lim_varepsilon rightarrow 0 fracP(A in a pm varepsilon cap B in b pm varepsilon) / varepsilonP(B in b pm varepsilon) / varepsilon $$

By definition, this is

$$frac f_A, B(a, b) f_B(b)$$

The answer given by Michael Hardy is correct. I just want to post a different interpretation of the same answer in case anyone finds it easier to follow:

I would like to compute the probability of $p(a=1|m=3)$:

$p(a=1|m=3) = fraca=1)p(a=1)p(m=3)$.

In this equation, $p(a=1)$ is a discrete probability which I can compute. The problem is computing $p(m=3)$ and $p(m=3|a=1)p(a=1)$ since $m$ is a continous random variable. Here is the solution:

For any contious variable, $f_X(x)$, we can write the probability of $x$ being equal to any constant c as follows:

$f_x(x=c) = f_x(c)dx$ which is the infinitely small area that amounts to the probability that we want to compute.

Applying this logic to above given equations, we:

$p(m=3|a=1) = f_m(m=3)dm p(a=1)$ and $f_m(m=3) = f_m(m=3)dm p(a=1) + f_m(m=3)dm p(a=0)$.

After putting everything together:

$p(a=1|m=3) = fraca=1)p(a=1)p(m=3) = fracf_m(m=3)dm p(a=1)f_m(m=3)dm p(a=1) + f_m(m=3)dm p(a=0) = fracf_m(m=3)p(a=1)f_m(m=3)p(a=1) + f_m(m=3)p(a=0)$

As seen this answer equals to the answer given by Michael Hardy.