# Hypothesis testing with the geometric distribution

A single observation of a random variable having a geometric distribution is used to test the null hypothesis $\theta=\theta_0$ against the alternative hypothesis $\theta=\theta_1 > \theta_0$.

If the null hypothesis is rejected if and only if the observed value of the random variable is greater than or equal to the positive integer $k$ , find expressions for the probabilities of type I and type II errors.

My idea:

Type I error equals $\alpha$, which is the probability that we reject $H_0$ while it is true. Probability of type II error is $\beta$, which is the probability that we accept $H_0$ while $H_0$ is false.

Let's say the Null Hypothetis is rejected, this means that $X\ge k$ ?

I know that that $P(X\ge k ; \theta_0)= 1- \theta_0(1-\theta_0)^{k-1}$. Is this value equal to $\alpha$?

How can I calculate $\beta$? $P(X\ge k ; \theta_1)= \theta_1(1-\theta_1)^{k-1}$.

Who can help me making this clear ? :-)

Thank you for your trouble :-)

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You have the definitions of Type I and Type II errors correct. However, the probability statement you use to represent the Type I error is incorrect. You want $P(X\geq k;\theta_0)=1-F_X(k-1;\theta_0)$, where $F_X(x;\theta_0)$ is the cumulative distribution function of a geometric random variable with parameter $\theta_0$. For the Type II error, the probability statement you've given is also incorrect. You want the probability of failing to reject the null hypothesis given that the alternative is true. This can be written as $P(X<k;\theta_1)=F_X(k-1;\theta_1)$.
When you write that $P(X\geq k;\theta_0)=1-\theta_0(1-\theta_0)^{k-1}$, you're actually saying that the Type I error is the probability of $X\neq k$ rather than $X\geq k$. Similarly, when you write that $P(X\geq k;\theta_1)=\theta_1(1-\theta_1)^{k-1}$, the inequality is incorrect (it should be $<$) and you've written that it is the probability of $X=k$ rather than $X<k$.