# Math Help - question on hypothesis

1. ## question on hypothesis

Under H0, a random variable has the cumulative distribution function $F_0(x)=x^2, 0\leq x\leq 1$; and under H1, it has the cumulative distribution function $F_1(x)=x^3, 0\leq x\leq 1$

Question: If the two hypotheses have equal prior probability, for what values of x is the posterior probability of H0 greater than that of H1?

This is what I did:

Since I know that $\frac{P(H_0|x)}{P(H_1|x)}=\frac{P(H_0)}{P(H_1)}\fr ac{P(x|H_0)}{P(x|H_1)}$, Given that the prior probability is equal will yield $\frac{P(H_0|x)}{P(H_1|x)}=\frac{P(x|H_0)}{P(x|H_1) }$ However, $x^2 \geq x^3$ given that 0<x<1, so it seems that for all values the statement is true, but the answer is apparent not as above.

Hope someone can help me. Thanks

2. The answer as given is 2/3. But I not sure how to get it

3. Originally Posted by noob mathematician
Under H0, a random variable has the cumulative distribution function $F_0(x)=x^2, 0\leq x\leq 1$; and under H1, it has the cumulative distribution function $F_1(x)=x^3, 0\leq x\leq 1$

Question: If the two hypotheses have equal prior probability, for what values of x is the posterior probability of H0 greater than that of H1?

This is what I did:

Since I know that $\frac{P(H_0|x)}{P(H_1|x)}=\frac{P(H_0)}{P(H_1)}\fr ac{P(x|H_0)}{P(x|H_1)}$, Given that the prior probability is equal will yield $\frac{P(H_0|x)}{P(H_1|x)}=\frac{P(x|H_0)}{P(x|H_1) }$ However, $x^2 \geq x^3$ given that 0<x<1, so it seems that for all values the statement is true, but the answer is apparent not as above.

Hope someone can help me. Thanks
$P(x|H_0)=\frac{d}{dx}x^2,\ \ 0

$P(x|H_1)=\frac{d}{dx}x^3, \ \ 0

CB

4. Given a level $\alpha$ test, I managed to find the rejection region as: $X>\sqrt{1-\alpha}$

What about the power of the test?

5. The answer is $1-(1-\alpha)^3 /2$. Anyone can help with the explanation?

My understanding is as followed: to find power of the test, I will use
$P(RejectH_0|H_1)$
= $P(X>\sqrt{1-\alpha}|H_1)$
= $1-\int_{0}^{\sqrt{1-\alpha}}3x^2.dx$
= $1-(1-\alpha)^{3/2}$

Can it be that the answer is wrong, which is highly impossible i think.