# question on hypothesis

• Oct 18th 2009, 04:56 AM
noob mathematician
question on hypothesis
Under H0, a random variable has the cumulative distribution function $\displaystyle F_0(x)=x^2, 0\leq x\leq 1$; and under H1, it has the cumulative distribution function $\displaystyle F_1(x)=x^3, 0\leq x\leq 1$

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 $\displaystyle \frac{P(H_0|x)}{P(H_1|x)}=\frac{P(H_0)}{P(H_1)}\fr ac{P(x|H_0)}{P(x|H_1)}$, Given that the prior probability is equal will yield $\displaystyle \frac{P(H_0|x)}{P(H_1|x)}=\frac{P(x|H_0)}{P(x|H_1) }$ However, $\displaystyle x^2 \geq x^3$ 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
• Oct 19th 2009, 04:02 AM
noob mathematician
The answer as given is 2/3. But I not sure how to get it
• Oct 19th 2009, 04:24 AM
CaptainBlack
Quote:

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

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 $\displaystyle \frac{P(H_0|x)}{P(H_1|x)}=\frac{P(H_0)}{P(H_1)}\fr ac{P(x|H_0)}{P(x|H_1)}$, Given that the prior probability is equal will yield $\displaystyle \frac{P(H_0|x)}{P(H_1|x)}=\frac{P(x|H_0)}{P(x|H_1) }$ However, $\displaystyle x^2 \geq x^3$ 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

$\displaystyle P(x|H_0)=\frac{d}{dx}x^2,\ \ 0<x<1$

$\displaystyle P(x|H_1)=\frac{d}{dx}x^3, \ \ 0<x<1$

CB
• Oct 19th 2009, 08:51 AM
noob mathematician
Given a level $\displaystyle \alpha$ test, I managed to find the rejection region as: $\displaystyle X>\sqrt{1-\alpha}$

What about the power of the test?
• Oct 20th 2009, 04:10 AM
noob mathematician
The answer is $\displaystyle 1-(1-\alpha)^3 /2$. Anyone can help with the explanation?

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

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