FB sample size题

Lets say the population on Facebook clicks ads with a

  click-through-rateof P. We select a sample of size N and examine the sample's conversion rate,d

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{P}is within DELTA of the true click through rate P, with 95% confidence.

网上有人回答的答案是这样的，觉得最后一个公式N is greater than (1 / delta)^2不是太懂，求讨论！

Interpret the question this way: we want to choose an N such that P_hat is an element of [P - delta, P + delta] with probability 95%.

First, note that since P_hat is the sum of N Bernoulli trials with some common parameter (by assumption) that we are trying to estimate, we can safely assume P_hat to be normally distributed with mean equal to the true mean (P) and variance equal to (P)(1 - P) / N.

Now, we when does a normally distributed random variable fall within delta of it's mean with 95% probability? The answer depends on how big delta is. Since P_hat is normally distributed, we know from our statistics classes that 95% of the time it will fall within 2 standard deviations of its mean.

So in other words, we want [P - delta, P + delta] = [P - 2*SE(P_hat), P + 2*SE(P_hat)]. That is, we want delta = SE(P_hat).

So what is the SE ("standard error") of P_hat? Well that's just the square root of its (sample) variance, or Sqrt(P_hat * (1 - P_hat) / N). But wait! We haven't run the experiment yet! How can we know what P_hat is?

We can either (a) make an educated guess, or (b) take the "worst" possible case and use that to upper bound N.
Let's go with option (b): P_hat * (1 - P_hat) is maximized when P_hat is .5, so the product is 0.25.

To put it all together: delta = 2 * Sqrt(0.25) / Sqrt(N) = 2 * .5 / Sqrt(N) => N = (1 / delta) ^ 2. So when N is greater than (1 / delta)^2, we can rest assured that P_hat will fall within the acceptable range 95% of the time.

因为这里要求minimum N, P_hat * (1 - P_hat) is maximized when P_hat is .5
so 2*sqrt(P_hat * (1 - P_hat))/sqrt(N) = delta >=2*0.5/sqrt(N)
then delta >=1/sqrt(N)
then sqrt(N)>=1/delta
so N>=(1/delta)^2
另外，这里应该是1.96 不是 2 吧？？

请问楼主面了吗？

cloud0325 发表于 2018-3-22 01:29
因为这里要求minimum N, P_hat * (1 - P_hat) is maximized when P_hat is .5
so 2*sqrt(P_hat * (1 - P_h ...

应该是1.96。楼主给的答案是用了3 sigma rule 估计的

Actually, it is not very important to use 1.96 or 2 since we use CLT to do the normal approximation.

Assume X_i = 1 if the i-th user click and 0 otherwise. Assume X_i ~iid Ber(p) and sum_{i=1}^N X_i ~ Bin(N,p). Let p_hat = \sum X_i / N be the sample click-through rate.
(1) E(p_hat) = E(sum X_i / N) = 1 / N sum E(X_i) = 1 / N * Np = p
(2) Var(p_hat) = Var(sum X_i / N) = 1 / N^2 sum Var(X_i) (by iid property)  = 1 / N^2 * Np(1-p) = p(1-p) / N.
By CLT, (p_hat - p) is asymptotically normally distributed with mean 0 and variance p(1-p)/N. That is, (p_hat - p) / sqrt(p(1-p) / N) ~ AN(0,1).

Note that
(1) 95% = P(|p_hat - p | < delta) =P(|(p_hat - p) / sqrt(p(1-p) / N) | < delta / sqrt(p(1-p) / N) ) and
(2) (p_hat - p) / sqrt(p(1-p) / N) ~ AN(0,1).
Hence, from the distribution function of N(0,1), we know that delta / sqrt(p(1-p) / N)  = 2 (or 1.96, whatever).

Now, rewrite the above formula gives N = 4 * delta^2 / p*(1-p). While we do not know the exact value of 0 < p < 1, it is clear that p(1-p) <= 1/4 by some simple calculus.

Finally, we get N =  4 * delta^2 / p*(1-p) >= 4 * delta^2 / 4 = delta^2.

h19881812 发表于 2018-3-25 23:59
Actually, it is not very important to use 1.96 or 2 since we use CLT to do the normal approximation. ...

Btw, when conducting a real experiment, we should be very careful about the assumption that X_i's are i.i.d.

h19881812 发表于 2018-3-25 23:59
Actually, it is not very important to use 1.96 or 2 since we use CLT to do the normal approximation. ...

很好的答案，， 解释的很清楚，但是你最后一步好像错了，，， N= 4*p(1-p) /belta^2    分子和分母反了~