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Assume we have a cross-section of $N$ stocks. $Y_i$ is an sample variance estimate of stock returns for stock $i$. This sample variance is estimated using $T_i$ number of observations. All $T_i$ are not necessarily equal, i.e. the sample size for $Y$ estimation differ for i = 1,2,.., N.

Now I want to run a cross-sectional weighted least squares regression:

$Y_i = \beta X_i + \epsilon_i$

What is the best choice of weights here, such that the weights are based on $T_i$ for each $Y_i$. In other words, I want to assign a smaller weight to stock $i$ if $T_i$ is small.

kjetil b halvorsen
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Mayou
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    If I understand it well, then you have $N$ stocks, and within each of these stocks you have observations $(x_{ij}, y_{ij})$, $i=1 \dots N, j=1 \dots n_i$ ? So I would suggest you to use the generalised least squares estimator. An R-implementation can be found in package nlme, function 'gls' where your grouping variable is the stock. You know R ? –  Nov 19 '15 at 07:14
  • As the the user before me wrote, look up [weighted least squares](https://en.wikipedia.org/wiki/Weighted_least_squares) and [generalized least squares](https://en.wikipedia.org/wiki/Generalized_least_squares). They are the canonical solutions to this problem, I believe. – jhin Aug 03 '21 at 21:33

3 Answers3

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I don't think there's a single optimal weight scheme here. I'd try first $w_i=\frac{NT_i}{\sum_iT_i}$. This way $\sum_iw_i=N$ and if $T_i=T_j\to w_i=1$, nice qualities.

Aksakal
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  • Thanks for your answer. Two follow-up questions please: 1) Do weights need to be normalized in weighted-least squares? 2) What do you think about $w_i = \sqrt{T_i}$ – Mayou Dec 05 '14 at 15:12
  • No, the weights don't have to be normalized, but it's nice to have them to be equal to 1 when num of obs are equal because then SSE will be the same as in OLS. It's just easier to compare and track the results. Square of $T_i$ is good too, because it links to random walk properties of volatility/time. – Aksakal Dec 05 '14 at 15:16
  • Great thank you. So in your opinion, what is the difference (advantage) of using $\sqrt{T_i}$ instead of $T_i$ as weights? – Mayou Dec 05 '14 at 15:28
  • Stocks with larger samples will have less impact than in linear weight – Aksakal Nov 11 '16 at 14:34
  • Unfortunately, these are the wrong weights to use, because the formula assumes *all* variation in the dependent values is due to measurement error. In effect, it assumes the regression model is perfect and has no error at all. If you had a prior sense of the variance $\sigma^2$ of the regression error, you could add it to each $N_i$ and use their inverses in a weighted regression. There are other issues to deal with, too, not least of which is the likely highly positively skewed distributions of the individual variance estimates. – whuber Feb 13 '22 at 17:38
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Yi (sample variance estimate of stock returns for stock i) is going to be too volatile. Replace it with a robust estimator like Median Absolute Deviation (M.A.D) in the weight function. I employed the latter successfully in a solvency model for insurance companies.

Also, if you regressed the sample variance estimate of stock returns against log (capitalization), a measure of a company's size, you should get an inverse smoothing effect as large companies have, on average, lower volatility in earnings and a lower stock beta. I would combine this with the M.A.D estimate.

AJKOER
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If my description of what you are doing is wrong please please correct me: We are supposed to have a set of valus {Xi} and the corresponding {Yi}. From simple least squares Y = A.X + B. Then we compute the total variance V = Σ(Yi - A.Xi - B)^2. It's a kind of iteration. Then we repeat minimization of the variance functional using the weights:

Wi = Vi / (Yi - A.Xi - B)^2

But then some of the Wi's may be infinite. I don't like this.

user143678
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  • Using math typesetting would make this easier to read. More information: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Sycorax Feb 13 '22 at 17:25
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    I don't like this, either, because I have tried this and it doesn't work, even when the weights converge. The problem is that the residuals reflect *both* the measurement error of the response variables *and* the error term in the regression, but this approach doesn't correctly separate the two. – whuber Feb 13 '22 at 17:31
  • Well, if we don't like results it seems the LS approximation falls foul somewhere further down the road, in some other calculation. In my case it's the Shannon predictivity parameter algorithm where I transport the LS data. Too small Shannon parameter means the LS approximation was not efficient. – user143678 Feb 14 '22 at 19:37