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

说，好的，对于每个 xi 我只执行这个操作。

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

我们要做的就在这里用 xi 的形式写下来。

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

因此，您获取总体中的每个数据点 - 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 mu 乘 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.

歌曲也翻译，但无论歌词否被理解，它们都会传达情感。

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 的，这太强烈了。

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

如果 xi 态分布的, 这种情况就会发生。

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

如果 习 不态分布的，这最终会发生。

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

所这 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.

所这看起来像一些 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，这将与 n 选择一些 习 成比。

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

所有xi的抽样分布什么?

Exactly, if all these xi's are independent.

确切地说, 如果所有这些xi都独立的。

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

但请记住, 习不独立的。

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

而所有的xi都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.

但这不仅仅一个通用的人工制品，我只需要知道 习 的总和。