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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?


1 Answer 1


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.

  • 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

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