主成分分析“向后”：给定的变量线性组合可解释多少数据差异？

我对六个变量$A$，$B$，$C$，$D$，$E$和进行了主成分分析$F$。如果我理解正确，未旋转的PC1会告诉我这些变量的线性组合描述/解释了数据中的最大方差，而PC2告诉我这些变量的线性组合描述了数据中的第二大方差，依此类推。

我只是很好奇-有什么办法可以做到这一点吗？假设我选择了这些变量的线性组合-例如$A+2B+5C$，我能算出所描述数据的方差是多少？

如果我们以所有变量都居中为前提（PCA中的标准做法），那么数据中的总方差就是平方和：

$$T=\sum_{i}(A_{i}^{2}+B_{i}^{2}+C_{i}^{2}+D_{i}^{2}+E_{i}^{2}+F_{i}^{2})$$

这等于变量的协方差矩阵的迹线，其等于协方差矩阵的特征值之和。这与PCA在“解释数据”方面所说的数量相同-即，您希望您的PC解释协方差矩阵的对角元素的最大比例。现在，如果我们将其作为一组预测值的目标函数，如下所示：

$$S=\sum_{i}\left(\left[A_{i}-\hat{A}_{i}\right]^{2}+\dots+\left[F_{i}-\hat{F}_{i}\right]^{2}\right)$$

然后，第一主成分最小化$S$所有秩1个拟合值之间$(\hat{A}_{i},\dots,\hat{F}_{i})$。所以看来您要追随的适当数量是

$$P=1-\frac{S}{T}$$ 要使用您的示例$A+2B+5C$，我们需要将此方程式转换为等级1的预测。首先，你需要标准化的权重有1平方之和所以我们更换$(1,2,5,0,0,0)$（平方和$30$用）$\left(\frac{1}{\sqrt{30}},\frac{2}{\sqrt{30}},\frac{5}{\sqrt{30}},0,0,0\right)$。接下来，我们根据归一化的权重对每个观察结果“打分”：

$$Z_{i}=\frac{1}{\sqrt{30}}A_{i}+\frac{2}{\sqrt{30}}B_{i}+\frac{5}{\sqrt{30}}C_{i}$$

然后，我们将分数乘以权重向量，得出排名1的预测。

$$\begin{pmatrix} \hat{A}_{i} \\ \hat{B}_{i} \\ \hat{C}_{i} \\ \hat{D}_{i} \\ \hat{E}_{i} \\ \hat{F}_{i}\end{pmatrix} =Z_{i}\times\begin{pmatrix} \frac{1}{\sqrt{30}} \\ \frac{2}{\sqrt{30}} \\ \frac{5}{\sqrt{30}} \\ 0 \\ 0 \\ 0\end{pmatrix}$$

Then we plug these estimates into $S$ calculate $P$. You can also put this into matrix norm notation, which may suggest a different generalisation. If we set $O$ as the $N\times q$ matrix of observed values of the variables ($q=6$ in your case), and $E$ as a corresponding matrix of predictions. We can define the proportion of variance explained as:

$$\frac{||O||_{2}^{2}-||O-E||_{2}^{2}}{||O||_{2}^{2}}$$

Where $||.||_{2}$ is the Frobenius matrix norm. So you could "generalise" this to be some other kind of matrix norm, and you will get a difference measure of "variation explained", although it won't be "variance" per se unless it is sum of squares.

Let's say I choose some linear combination of these variables -- e.g. $A+2B+5C$, could I work out how much variance in the data this describes?

This question can be understood in two different ways, leading to two different answers.

A linear combination corresponds to a vector, which in your example is $[1, 2, 5, 0, 0, 0]$. This vector, in turn, defines an axis in the 6D space of the original variables. What you are asking is, how much variance does projection on this axis "describe"? The answer is given via the notion of "reconstruction" of original data from this projection, and measuring the reconstruction error (see Wikipedia on Fraction of variance unexplained). Turns out, this reconstruction can be reasonably done in two different ways, yielding two different answers.

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Approach #1

Let $\newcommand{\S}{\boldsymbol \Sigma} \newcommand{\w}{\mathbf w} \newcommand{\v}{\mathbf v}\newcommand{\X}{\mathbf X} \X$ be the centered dataset ($n$ rows correspond to samples, $d$ columns correspond to variables), let $\S$ be its covariance matrix, and let $\w$ be a unit vector from $\mathbb R^d$. The total variance of the dataset is the sum of all $d$ variances, i.e. the trace of the covariance matrix: $T = \mathrm{tr}(\S)$. The question is: what proportion of $T$ does $\w$ describe? The two answers given by @todddeluca and @probabilityislogic are both equivalent to the following: compute projection $\X \w$, compute its variance and divide by $T$:

$$R^2_\mathrm{first} = \frac{\mathrm{Var}(\X \w)}{T} = \frac{\w^\top \S \w}{\mathrm{tr}(\S)}.$$

This might not be immediately obvious, because e.g. @probabilityislogic suggests to consider the reconstruction $\X \w \w^\top$ and then to compute

$$\frac{\|\X\|^2 - \|\X-\X \w \w^\top\|^2}{\|\X\|^2},$$ but with a little algebra this can be shown to be an equivalent expression.

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Approach #2

Okay. Now consider a following example: $\X$ is a $d=2$ dataset with covariance matrix

$$\S = \left(\begin{array}{c}1&0.99\\0.99&1\end{array}\right)$$ and $\mathbf w = (\begin{array}{}1&0\end{array})^\top$ is simply an $x$ vector:

The total variance is $T=2$. The variance of the projection onto $\w$ (shown in red dots) is equal to $1$. So according to the above logic, the explained variance is equal to $1/2$. And in some sense it is: red dots ("reconstruction") are far away from the corresponding blue dots, so a lot of the variance is "lost".

On the other hand, the two variables have $0.99$ correlation and so are almost identical; saying that one of them describes only $50\%$ of the total variance is weird, because each of them contains "almost all the information" about the second one. We can formalize it as follows: given projection $\X\w$, find a best possible reconstruction $\X\w\v^\top$ with $\v$ not necessarily the same as $\w$, and then compute the reconstruction error and plug it into the expression for the proportion of explained variance:

$$R^2_\mathrm{second}=\frac{\|\X\|^2 - \|\X-\X \w \v^\top\|^2}{\|\X\|^2},$$ where $\v$ is chosen such that $\|\X-\X \w \v^\top\|^2$ is minimal (i.e. $R^2$ is maximal). This is exactly equivalent to computing $R^2$ of multivariate regression predicting original dataset $\X$ from the $1$-dimensional projection $\X\w$.

It is a matter of straightforward algebra to use regression solution for $\v$ to find that the whole expression simplifies to

$$R^2_\mathrm{second}=\frac{\|\S \w\|^2}{\w^\top \S \w \cdot \mathrm{tr}(\S)}.$$ In the example above this is equal to $0.9901$, which seems reasonable.

Note that if (and only if) $\w$ is one of the eigenvectors of $\S$, i.e. one of the principal axes, with eigenvalue $\lambda$ (so that $\S \w = \lambda \w$), then both approaches to compute $R^2$ coincide and reduce to the familiar PCA expression

$$R^2_\mathrm{PCA} = R^2_\mathrm{first} = R^2_\mathrm{second} = \lambda/\mathrm{tr}(\S) = \lambda/\sum \lambda_i.$$

PS. See my answer here for an application of the derived formula to the special case of $\w$ being one of the basis vectors: Variance of the data explained by a single variable.

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Appendix. Derivation of the formula for $R^2_\mathrm{second}$

Finding $\v$ minimizing the reconstruction $\|\X-\X \w \v^\top\|^2$ is a regression problem (with $\X \w$ as univariate predictor and $\X$ as multivariate response). Its solution is given by

$$\v^\top = \left((\X \w)^\top (\X \w)\right)^{-1}(\X \w)^\top \X = (\w^\top \S \w)^{-1} \w^\top \S.$$

Next, the $R^2$ formula can be simplified as

$$R^2=\frac{\|\X\|^2 - \|\X-\X \w \v^\top\|^2}{\|\X\|^2} = \frac{\|\X \w \v^\top\|^2}{\|\X\|^2}$$ due to the Pythagoras theorem, because the hat matrix in regression is an orthogonal projection (but it is also easy to show directly).

Plugging now the equation for $\v$, we obtain for the numerator:

$$\|\X \w \v^\top\|^2 = \mathrm{tr}\left(\X \w \v^\top (\X \w \v^\top)^\top\right) = \mathrm{tr}(\X\w\w^\top\S\S\w\w^\top\X^\top)/(\w^\top\S\w)^2=\mathrm{tr}(\w^\top\S\S\w)/(\w^\top\S\w) = \|\S\w\|^2 / (\w^\top\S\w).$$

The denominator is equal to $\|\X\|^2 = \mathrm{tr}(\S)$ resulting in the formula given above.

Let the total variance, $T$, in a data set of vectors be the sum of squared errors (SSE) between the vectors in the data set and the mean vector of the data set,

$$T = \sum_{i} (x_i-\bar{x}) \cdot (x_i-\bar{x})$$ where $\bar{x}$ is the mean vector of the data set, $x_i$ is the ith vector in the data set, and $\cdot$ is the dot product of two vectors. Said another way, the total variance is the SSE between each $x_i$ and its predicted value, $f(x_i)$, when we set $f(x_i)=\bar{x}$.

Now let the predictor of $x_i$, $f(x_i)$, be the projection of vector $x_i$ onto a unit vector $c$.

$$f_c(x_i) = (c \cdot x_i)c$$

Then the $SSE$ for a given $c$ is

$$SSE_c = \sum_i (x_i - f_c(x_i)) \cdot (x_i - f_c(x_i))$$

I think that if you choose $c$ to minimize $SSE_c$, then $c$ is the first principal component.

If instead you choose $c$ to be the normalized version of the vector $(1, 2, 5, ...)$, then $T-SSE_c$ is the variance in the data described by using $c$ as a predictor.