You can just say, OK, for each xi I just perform this operation.

你可以说，好的，对于每 xi 我只执行作。

And what we want to do is write this right here just in terms of xi's.

我们要做的就是在里用 xi 的形式写下来。

So you take each of the data points in the population-- xi.

因此，您获取总体中的每数据点 - xi。

I plug in the actual data, the XIs.

我入实际数据，即 XI。

Edit, Undo, minus 2 times mu times the sum from i is equal to 1 to n of xi.

编辑，撤消，减去 2 乘以 mu 乘以 i 的和等于 xi 的 1 到 n。

The songs can also be translated, but they communicate emotionally whether xi or not the words are understood.

歌曲也可以翻译，但无论歌词是否被理解，它们都会感。

And it turns out my problem set up, the XIs are normal is too strong of a statement.

事实证明，我的问题设置，XI 是正常的，太强烈了。

If the xi's were normally distributed, this just happens.

如果 xi 是正态分布的, 种况就会发生。

If the xi's are not normally distributed, this eventually happens.

如果 习 不是正态分布的，最终会发生。

So this is p to the sum of the xi's plus alpha minus 1.

所以是 p 与 xi 的总和加上 alpha 减去 1。

1 Minus p to the n minus sum of the xi's plus beta minus 1.

1减去p到n减去xi的和加上β减去1。

We see some xi's from a Bernoulli distribution.

我们从伯努利分布中看到了一些xi。

So this looks like some of the xi's plus alpha over n plus alpha plus beta.

所以看起来像一些 习 加 alpha 超过 n 加 alpha 加 beta。

For p, given all of my data, this is going to be proportional to n choose some of the xi's.

对于 p，给定我的所有数据，将与 n 选择一些 习 成正比。

What's the sampling distribution for all of the xi's?

所有xi的抽样分布是什么?

Exactly, if all these xi's are independent.

确切地说, 如果所有些xi都是独立的。

But keep in mind, the xi's are not independent.

但请记住, 习不是独立的。

And all of the xi's is n-dimensional.

而所有的xi都是n维的。

The sum of the xi's is sufficient.

xi 的总和就足够了。

But it's more than just a general artifact that all I need to know is the sum of the xi's.

但不仅仅是一通用的人工制品，我只需要知道 习 的总和。