0
$\begingroup$

The exercise

We have $\Theta = \{0,1\}$ and let $X$ be a random variable with density function $f(x;0)=1$ and $f(x;1)=3x^2$ for $x\in (0,1)$. I want to find the most powerful test of size $\alpha=0.2$ for the testing problem: $H_0: \theta=0$ vs $H_1: \theta=1.$

I have done a similar exercise for a given discrete distribution, but this one is quite different.

What I have done so far

My first thoughts were to let $X_1,...,X_n$ be a random sample from $X$, and then search for a critical region $$R=\left\{\underset{\sim}{X}: \frac{f_1(\underset{\sim}{X})}{f_0(\underset{\sim}{X})} > k\right\} ,$$ and then choose $k$ such that $P_{\theta_0}(\underset{\sim}{X} \in R)=\alpha.$

The ratio can then be calculated as

$$\frac{f_1(\underset{\sim}{X})}{f_0(\underset{\sim}{X})} = \frac{f_1(\underset{\sim}{X};1)}{f_0(\underset{\sim}{X};0)} = 3 \sum_{i=1}^n X_i^2,$$

and therefore

$$P_{\theta_0}\left\{ \frac{f_1(\underset{\sim}{X})}{f_0(\underset{\sim}{X})}>k\right\} =P_{\theta_0}\left\{3 \sum_{i=1}^n X_i^2 >k\right\}.$$

This is where I become stuck, so I would really appreciate some feedback on a) if I have done the right stuff so far, and b) How I should now proceed.

$\endgroup$
6
  • $\begingroup$ How do you get a sum in $X_i^2$ when evaluating $f_1$? Can you elaborate the steps there? $\endgroup$
    – Glen_b
    Commented Jan 12, 2022 at 22:49
  • $\begingroup$ I was thinking since we had many random samples, but I now believe it becomes sufficient to use just one random sample, which in this case gives a final $k=1.92$. Does that make sense? @Glen_b $\endgroup$
    – Joe
    Commented Jan 13, 2022 at 8:28
  • $\begingroup$ I'm sorry but you have missed the point. Keep all $n$ points in the sample. How did you get a SUM there: $\frac{f_1(\underset{\sim}{X};1)}{f_0(\underset{\sim}{X};0)} = 3 \sum_{i=1}^n X_i^2,$ You very much need to justify that step. $\endgroup$
    – Glen_b
    Commented Jan 14, 2022 at 3:26
  • $\begingroup$ Ah it should be a product of $X_i$s... $\endgroup$
    – Joe
    Commented Jan 14, 2022 at 9:02
  • $\begingroup$ Be careful, though, that's not the only thing to worry about there. $\endgroup$
    – Glen_b
    Commented Jan 14, 2022 at 12:51

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.