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Math 135B: HW # 1
1. Statistical Hypothesis Testing: In class we showed that if Sn = X1 + · · · + Xn
where Xj are independent and identically distributed Bernoulli random variables and 1 > a > p > 0
where p = E(Xj ), then1 P(Sn ≥ na) ≈ e −nΛ∗(a,p)
where Λ ∗ (a,p) = a log(a p ) + b log( b q ) where q = 1 − p and b = 1 − a.
Suppose now that p is unknown and that there are two competing hypotheses concerning its value, the hypothesis H1 that p = p1 and the hypothesis H2 that p = p2 where p1 < p2.
Given the observed results X1, . . . , Xn of n Bernoulli trials, one decides in favor of H2 if Sn ≥ na and in favor of H1 if Sn < na,
where a is some number satisfying p1 < a < p2.
The question is, how do we choose a? Her is one strategy. Given the hypothesis H1: p = p1, then P(Sn ≥ na|H1) ≈ e −nΛ∗(a,p1) whereas given the hypothesis H2: p = p2, P(Sn < na|H2) ≈ e −nΛ∗(a,p2) These give us the probabilities of erroneously deciding in favor of H2 whe H1 is, in fact, true and of erroneously deciding in favor of H1 when H2 is in fact true.
Since we don’t want to favor one event over another we choose a so that the two error probabilities are approximately equal, e.g. Λ ∗ (a,p1) = Λ∗ (a,p2).
This gives an equation for the unknown a.
Show that the a that satifies this equation is
a = log(q1/q2) log(p2/p1) + log(q1/q2) 1 If a < p then a similar argument shows that P(Sn < na) ≈ e −nΛ ∗(a,p)
Let p2 = p1 + t. t > 0. Show that for small t a = p + 1 2 t + p − q 12pq t 2 + 1 − 2p + 2p 2 24q 2p 2 t 3 + O(t 4 )
and that
Λ ∗ (a,p) = t 2 8pq + p − q 16p 2q 2 t 3 + 23 − 65p + 65p 2 576q 3p 3 t 4 + O(t 5 ).
You might want to use Mathematica to perform these last two computations.
2. Grimmett & Stirzaker: page 219, #1.
3. Grimmett & Stirzaker: page 219, #2.