A basketball probability question using Neyman–Pearson lemma It is known that the probability of a basketball player to make his first shot is $p=0.6$
A player argues that it does not matter if he made the previous shot or not his odds stays the same. We say if he misses the previous shot his odds go down.
In order to check out our theory the player was asked to throw shots until his first success or $5$ consecutive fails. The test will only take place if he misses the first shot.
Let $X_i$ be an indicator whether he made the $i$ shot or not($x_i=1$ successful shot,$x_i =0$ fail)
Let Y be the number of the first shot that he made successfully
Define:
$H_0 : P(X_i =1|X_1=0,\ldots,X_{i-1}=0) = p \qquad \qquad i=2,\ldots,Y $
$H_1 : P(X_i =1|X_1=0,\ldots,X_{i-1}=0) = rp \qquad \qquad r < 1, \quad i=2,\ldots,Y $
I need to use Neyman–Pearson lemma to build a UMP test with Statistical significance of $\alpha =0.1$
Usually statistics is my strong side but this area does not fill very well to me, i tried several ways but all came to results that does not seem right
What is the right way to approach this question? and how can i find a UMP test for it?
 A: To construct a Neyman-Pearson test, for $H_0:\theta=p$ vs $H_1:\theta=pr$ we need to find a critical region of the form
$$\frac{L(p;n;\boldsymbol{x})}{L(pr;n;\boldsymbol{x})} \leq A$$
Where $L(\theta;n;\boldsymbol{x})$ is the likelihood function (given n throws were made).  Since we're dealing with independent Bernoulli observations (and the first throw has probability $p$
$$L(\theta;\boldsymbol{x}) = \big(px_1+(1-p)(1-x_1))\prod_{i=2}^{n}\big((\theta x_i +(1-\theta)(1-x_i)\big) $$
(Sorry this is probably a little sloppy notation wise, so I'm open to comment for how to rephrase...it shouldn't matter too much).
Now, to find $A$, we require that
$$\alpha = P\Big(\frac{L(p;n;x)}{L(pr;n;x)}\leq A|H_0\Big)$$
Note that the right hand probability is just the sum of $P(n \text{ shots made}|H_0)$ where $n$ (and corresponding $x$) results in $\frac{L(p;n;x)}{L(pr;n;x)}\leq A$, and you just chose $A$ so that the sum is $\leq \alpha$
Note:  $P(n \text{ shots made}|H_0)$ for n=5 will need to be divided into the cases where the shot is and is not made (because the game automatically ends after 5 shots) ...
