有效预测用户点击超链接的可能性
可能的重复：
确定用户点击超链接的概率 < /p>
我的网页上有一堆超链接。 根据过去的观察，我知道用户点击每个超链接的概率。 因此，我可以计算这些概率的平均值和标准差。
我现在添加一个新的超链接到此页面。 经过短暂的测试后，我发现在看到此超链接的 20 个用户中，有 5 个用户点击了它。
考虑到其他超链接的点击概率的已知平均值和标准差（这形成“先验期望”），如何有效地估计用户点击新超链接的概率？
一个天真的解决方案是忽略其他概率，在这种情况下我的估计只是 5/20 或 0.25 - 但这意味着我们正在丢弃相关信息，即我们之前对点击概率的期望。
所以我正在寻找一个看起来像这样的函数：
double estimate(double priorMean, double priorStandardDeviation, int clicks, int views);
我会问，因为我比数学符号更熟悉代码，所以任何答案都优先使用代码或伪代码而不是数学。
Possible Duplicate:
Determining the probability of a user clicking a hyperlink
So I have a bunch of hyperlinks on a web page. From past observation I know the probabilities that a user will click on each of these hyperlinks. I can therefore calculate the mean and standard deviation of these probabilities.
I now add a new hyperlink to this page. After a short amount of testing I find that of the 20 users that see this hyperlink, 5 click on it.
Taking into account the known mean and standard deviation of the click-through probabilities on other hyperlinks (this forms a "prior expectation"), how can I efficiently estimate the probability of a user clicking on the new hyperlink?
A naive solution would be to ignore the other probabilities, in which case my estimate is just 5/20 or 0.25 - however this means we are throwing away relevant information, namely our prior expectation of what the click-through probability is.
So I'm looking for a function that looks something like this:
double estimate(double priorMean, double priorStandardDeviation, int clicks, int views);
I'd ask that, since I'm more familiar with code than mathematical notation, that any answers use code or pseudocode in preference to math.
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我不想在这里给出一个非答案，但是您的用户单击链接的可能性不取决于链接的类型吗？ 或者这是某种广告？ 即便如此，它仍然取决于内容，不是吗？ 我很想说这更像是一个心理学问题而不是统计学问题......;]
I hate to give a non-answer here, but doesn't the likelihood that your user click the link depend on the type of link it is? Or is this ads of some sort? Even then, it would be content-dependent, wouldn't it? I would be tempted to say this is more of a psychology question than a statistics one... ;]