I'm trying to solve a problem for least angle regression (LAR). This is a problem 3.23 on page 97 of Hastie et al., Elements of Statistical Learning, 2nd. ed. (5th printing).

Consider a regression problem with all variables and response having mean zero and standard deviation one. Suppose also that each variable has identical absolute correlation with the response:

$ \frac{1}{N} | \left \langle \bf{x}_j, \bf{y} \right \rangle | = \lambda, j = 1, ..., p $

Let $\hat{\beta}$ be the least squares coefficient of $\mathbf{y}$ on $\mathbf{X}$ and let $\mathbf{u}(\alpha)=\alpha \bf{X} \hat{\beta}$ for $\alpha\in[0,1]$.

I am asked to show that $$ \frac{1}{N} | \left \langle \bf{x}_j, \bf{y}-u(\alpha) \right \rangle | = (1 - \alpha) \lambda, j = 1, ..., p $$ and I am having problems with that. Note that this can basically says that the correlations of each $x_j$ with the residuals remain equal in magnitude as we progress toward $u$.

I also do not know how to show that the correlations are equal to:

$\lambda(\alpha) = \frac{(1-\alpha)}{\sqrt{(1-\alpha)^2 + \frac{\alpha (2-\alpha)}{N} \cdot RSS}} \cdot \lambda$

Any pointers would be greatly appreciated!

  • 2
    $\begingroup$ @Belmont, what is $u(\alpha)$? Could you provide more context about your problem? Link to article with standard properties of LAR for example would help a lot. $\endgroup$
    – mpiktas
    Feb 2, 2011 at 14:00
  • $\begingroup$ @Belmont, This looks like a problem from Hastie, et al., Elements of Statistical Learning, 2nd. ed. Is this homework? If so, you might add that tag. $\endgroup$
    – cardinal
    Feb 3, 2011 at 1:50
  • $\begingroup$ @Belmont, now that @cardinal gave a complete answer, can you specify what LAR really is, for future reference? Judging from the answer this is standard manipulation of products of least squares regressions given some initial constraints. There should not be a special name for it without serious reason. $\endgroup$
    – mpiktas
    Feb 8, 2011 at 7:59
  • 1
    $\begingroup$ @mpiktas, it's a stagewise algorithm, so each time a variable enters or leaves the model on the regularization path, the size (i.e., cardinality/dimension) of $\beta$ grows or shrinks respectively and a "new" LS estimate is used based on the currently "active" variables. In the case of the lasso, which is a convex optimization problem, the procedure is is essentially exploiting special structure in the KKT conditions to obtain a very efficient solution. There are also generalizations to, e.g., logistic regression based on IRLS and Heine-Borel (to prove convergence in finite no. of steps.) $\endgroup$
    – cardinal
    Feb 8, 2011 at 13:19
  • 1
    $\begingroup$ @Belmont -1, as I recently bought the book of Hastie, I can confirm, that this is an exercise from it. So I am giving you a big -1, since you do not even manage to give all the definitions, I am not even talking about giving the reference. $\endgroup$
    – mpiktas
    Feb 9, 2011 at 14:14

1 Answer 1


This is problem 3.23 on page 97 of Hastie et al., Elements of Statistical Learning, 2nd. ed. (5th printing).

The key to this problem is a good understanding of ordinary least squares (i.e., linear regression), particularly the orthogonality of the fitted values and the residuals.

Orthogonality lemma: Let $X$ be the $n \times p$ design matrix, $y$ the response vector and $\beta$ the (true) parameters. Assuming $X$ is full-rank (which we will throughout), the OLS estimates of $\beta$ are $\hat{\beta} = (X^T X)^{-1} X^T y$. The fitted values are $\hat{y} = X (X^T X)^{-1} X^T y$. Then $\langle \hat{y}, y-\hat{y} \rangle = \hat{y}^T (y - \hat{y}) = 0$. That is, the fitted values are orthogonal to the residuals. This follows since $X^T (y - \hat{y}) = X^T y - X^T X (X^T X)^{-1} X^T y = X^T y - X^T y = 0$.

Now, let $x_j$ be a column vector such that $x_j$ is the $j$th column of $X$. The assumed conditions are:

  • $\frac{1}{N} \langle x_j, x_j \rangle = 1$ for each $j$, $\frac{1}{N} \langle y, y \rangle = 1$,
  • $\frac{1}{N} \langle x_j, 1_p \rangle = \frac{1}{N} \langle y, 1_p \rangle = 0$ where $1_p$ denotes a vector of ones of length $p$, and
  • $\frac{1}{N} | \langle x_j, y \rangle | = \lambda$ for all $j$.

Note that in particular, the last statement of the orthogonality lemma is identical to $\langle x_j, y - \hat{y} \rangle = 0$ for all $j$.

The correlations are tied

Now, $u(\alpha) = \alpha X \hat{\beta} = \alpha \hat{y}$. So, $$ \langle x_j, y - u(a) \rangle = \langle x_j, (1-\alpha) y + \alpha y - \alpha \hat{y} \rangle = (1-\alpha) \langle x_j, y \rangle + \alpha \langle x_j, y - \hat{y} \rangle , $$ and the second term on the right-hand side is zero by the orthogonality lemma, so $$ \frac{1}{N} | \langle x_j, y - u(\alpha) \rangle | = (1-\alpha) \lambda , $$ as desired. The absolute value of the correlations are just $$ \hat{\rho}_j(\alpha) = \frac{\frac{1}{N} | \langle x_j, y - u(\alpha) \rangle |}{\sqrt{\frac{1}{N} \langle x_j, x_j \rangle }\sqrt{\frac{1}{N} \langle y - u(\alpha), y - u(\alpha) \rangle }} = \frac{(1-\alpha)\lambda}{\sqrt{\frac{1}{N} \langle y - u(\alpha), y - u(\alpha) \rangle }} $$

Note: The right-hand side above is independent of $j$ and the numerator is just the same as the covariance since we've assumed that all the $x_j$'s and $y$ are centered (so, in particular, no subtraction of the mean is necessary).

What's the point? As $\alpha$ increases the response vector is modified so that it inches its way toward that of the (restricted!) least-squares solution obtained from incorporating only the first $p$ parameters in the model. This simultaneously modifies the estimated parameters since they are simple inner products of the predictors with the (modified) response vector. The modification takes a special form though. It keeps the (magnitude of) the correlations between the predictors and the modified response the same throughout the process (even though the value of the correlation is changing). Think about what this is doing geometrically and you'll understand the name of the procedure!

Explicit form of the (absolute) correlation

Let's focus on the term in the denominator, since the numerator is already in the required form. We have $$ \langle y - u(\alpha), y - u(\alpha) \rangle = \langle (1-\alpha) y + \alpha y - u(\alpha), (1-\alpha) y + \alpha y - u(\alpha) \rangle . $$

Substituting in $u(\alpha) = \alpha \hat{y}$ and using the linearity of the inner product, we get

$$ \langle y - u(\alpha), y - u(\alpha) \rangle = (1-\alpha)^2 \langle y, y \rangle + 2\alpha(1-\alpha) \langle y, y - \hat{y} \rangle + \alpha^2 \langle y-\hat{y}, y-\hat{y} \rangle . $$

Observe that

  • $\langle y, y \rangle = N$ by assumption,
  • $\langle y, y - \hat{y} \rangle = \langle y - \hat{y}, y - \hat{y} \rangle + \langle \hat{y}, y - \hat{y} \rangle = \langle y - \hat{y}, y - \hat{y}\rangle$, by applying the orthogonality lemma (yet again) to the second term in the middle; and,
  • $\langle y - \hat{y}, y - \hat{y} \rangle = \mathrm{RSS}$ by definition.

Putting this all together, you'll notice that we get

$$ \hat{\rho}_j(\alpha) = \frac{(1-\alpha) \lambda}{\sqrt{ (1-\alpha)^2 + \frac{\alpha(2-\alpha)}{N} \mathrm{RSS}}} = \frac{(1-\alpha) \lambda}{\sqrt{ (1-\alpha)^2 (1 - \frac{\mathrm{RSS}}{N}) + \frac{1}{N} \mathrm{RSS}}} $$

To wrap things up, $1 - \frac{\mathrm{RSS}}{N} = \frac{1}{N} (\langle y, y, \rangle - \langle y - \hat{y}, y - \hat{y} \rangle ) \geq 0$ and so it's clear that $\hat{\rho}_j(\alpha)$ is monotonically decreasing in $\alpha$ and $\hat{\rho}_j(\alpha) \downarrow 0$ as $\alpha \uparrow 1$.

Epilogue: Concentrate on the ideas here. There is really only one. The orthogonality lemma does almost all the work for us. The rest is just algebra, notation, and the ability to put these last two to work.

  • 2
    $\begingroup$ @cardinal, +1. The answer is magnitudes better than the question. $\endgroup$
    – mpiktas
    Feb 8, 2011 at 7:56
  • $\begingroup$ @cardinal, you might want to change the link to amazon or some other site. I think that linking to the full book might raise some copyright issues. $\endgroup$
    – mpiktas
    Feb 9, 2011 at 14:09
  • 3
    $\begingroup$ @mpiktas, nope. No copyright issues. That is the official website for the book. The authors obtained permission from Springer to make the PDF freely available online. (See the note to this effect on the site.) I think they got the idea from Stephen Boyd and his Convex Optimization text. Hopefully such a trend will pick up steam over the next few years. Enjoy! $\endgroup$
    – cardinal
    Feb 9, 2011 at 14:30
  • $\begingroup$ @cardinal, ooh massive thanks! That is mighty generous from the authors. $\endgroup$
    – mpiktas
    Feb 9, 2011 at 14:42
  • $\begingroup$ @mpiktas, it's by far the most popular book in the Springer Series in Statistics. It looks good on an iPad. Which reminds me---I should download Boyd's text onto it as well. Cheers. $\endgroup$
    – cardinal
    Feb 9, 2011 at 14:47

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