4
$\begingroup$

enter image description here

In this question, I tried using the fundamental property of the Wald Sequential Probability Ratio Test (SPRT): $P( N = \infty )=0.$

For this I calculated $p_1 = 1- P(-1.5 < S_1 < 2.5 )$, that is probability of termination of procedure in first trial, and $p_1 = 1- \theta - (1-\theta) = 0$. Furthermore, also $p_2, p_3$ came out to be 0.

As this contradicts Wald's fundamental property, the procedure shouldn't be an SPRT.

Am I correct?

$\endgroup$

1 Answer 1

4
+50
$\begingroup$

Following the wikipedia page and this presentation:

$$H_0: \theta = \theta_0 = \frac{1}{4}$$

$$H_1: \theta = \theta_1 = \frac{3}{4}$$

$$A \sim \ln\frac{\beta}{1-\alpha}$$

$$B \sim \ln\frac{1-\beta}{\alpha}$$

$$\Lambda_i = \ln\frac{L(\theta_1|x_i)}{L(\theta_0|x_i)}.$$

Accept $H_0$ if $\sum \Lambda_i < A$, Accept $H_1$ if $\sum \Lambda_i > B$, and continue sampling if $A < \sum \Lambda_i < B. $

Assume $\beta = \alpha = 0.0122 = \frac{3^4-1}{3^8-1}$ (I know this works in the forward direction, but you can also solve backward from the problem statement)

$$\ln\frac{\beta}{1-\alpha} < \sum_1^n \ln \frac{\theta_1^{x_i}(1-\theta_1)^{1-x_i}}{\theta_0^{x_i}(1-\theta_0)^{1-x_i}} < \ln\frac{1-\beta}{\alpha}$$

$$\ln(1/3^4) < \ln\left(\frac{\theta_1}{\theta_0}\right) \sum x_i + \ln\left(\frac{1-\theta_1}{1-\theta_0}\right) \sum (1-x_i) < \ln(3^4)$$

$$-4\ln3 < \ln3 \sum x_i + \ln\frac{1}{3} \sum (1-x_i) < 4\ln3$$

$$-4\ln3 < \ln3 \sum x_i - \ln3 (n - \sum x_i) < 4\ln3$$

$$-4\ln3 < 2\ln3 \sum x_i - n\ln3 < 4\ln3$$

$$n\ln3-4\ln3 < 2\ln3 \sum x_i < n\ln3 + 4\ln3$$

$$n/2 - 2 < \sum x_i < n/2 + 2$$

Therefore, this test is an SPRT.

$\endgroup$
3
  • 1
    $\begingroup$ Concise post. +1. $\endgroup$ Commented Oct 16, 2022 at 2:50
  • $\begingroup$ You assumed $ \alpha , \beta $ as 0.0122 , what is the reason behind that, or do we have to assume this value only in every such question? $\endgroup$
    – simran
    Commented Oct 16, 2022 at 2:59
  • 1
    $\begingroup$ @simran I figured out the value of $\alpha$ and $\beta$ after I got to the end of the proof. I worked backward to figure out $\alpha$ and $\beta$ I needed to get the exact answer from the original question, specifically the $\pm2$. For ease of showing the steps I just started with the right $\alpha$ and $\beta$. If you keep $\alpha$ and $\beta$ in your proof, you will see what I mean when you get to the end. $\endgroup$
    – R Carnell
    Commented Oct 16, 2022 at 19:35

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.