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I have read some papers about Weighted Distribution. Suppose $X$ is a non-negative continuous random variable with pdf $f(x)$. The pdf of the weighted random variable $X_w$ is given by:

$f_w (x) = \frac{w(x)f(x)}{E[w(X)]}$, where $w(x)$ is a non-negative weight function.

My question is: why do we need to add weight function $w(x)$ to the $f(x)$? I mean, what is the idea behind the weighted distribution? What are the effects after adding the $w(x)$ to the $f(x)$? I still can't find the answer that I want.

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    $\begingroup$ (What do you think would happen / why do you think it would be useful) without the weights? $\endgroup$ Nov 1, 2021 at 15:06
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    $\begingroup$ Why do you requite $X$ no-negative? It is unnecessary!, but you maybe need $w(x)$ non-negative ... $\endgroup$ Nov 1, 2021 at 15:49
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    $\begingroup$ $w$ is the Radon-Nikodym derivative of $f_w$ with respect to $f.$ See stats.stackexchange.com/questions/326185/…. For applications, search our site. Looking up "change of measure" in financial applications could be of interest, too. $\endgroup$
    – whuber
    Nov 1, 2021 at 16:36

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Situations in which weighted distributions occur or have some use:

  1. Mixture models of the type $f(x)=\sum_{k=1}^K \pi_kf_k(x)$. To clarify: not the mixture itself is a weighted distribution, rather a mixture component $f_k$ is $f$ weighted by $w(x)$ being the probability $p_k$ that $x$ has been generated by mixture component $f_k$. Formally: $f_k(x)=f(x|Z=k)=\frac{p_k(x)f(x)}{Ep_k(x)}$, where $Z$ is a random variable giving the component memberships with probabilities $\pi_1,\ldots,\pi_K$ (in fact $Ep_k(x)=\pi_k$). This can be used in an algorithm (EM) to iteratively estimate the parameters of $f_k$ given the observation weights, and then the observation weights given the estimated parameters.

  2. Some methods for identifying outliers and robust estimators estimate robustness weights (sometimes but not necessarily interpreted as probabilities that observations are not outliers). One may be interested in the distribution of non-outliers, which would be the overall distribution weighted by robustness weights.

More generally one may be interested in a subpopulation of the data without having "hard" information about which observations belong to this subpopulation, and either known or estimated weights (once more these can be probabilities but don't necessarily have to) that specify the degree to which the observations belong to the subpopulation of interest.

There's also an application in sampling theory (although I'm not an expert for this). If you want to represent an underlying population, but you have more observations than proportionally required of one part of it and less of another part, you have observed a weighted form of the original distribution with higher weights for parts that are more likely to be in your sample (and you may want to downweight these when estimating the underlying population distribution).

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  • $\begingroup$ I think you might be discussing a different concept: the mixture is not a weighted distribution in the sense defined in the question. $\endgroup$
    – whuber
    Nov 1, 2021 at 15:39
  • $\begingroup$ @whuber Not the mixture itself is a weighted distribution, but weighted distributions occur when estimating the individual mixture components $f_k$ using the EM-algorithm (and also in some theory investigating such estimators). $\endgroup$ Nov 1, 2021 at 15:44
  • $\begingroup$ @ChristianHennig Thank you very much for your answer. I still have a question that I want to ask. I read in the paper and it is said that the weighted distributions refer to instances that the recorded observations can't be considered as a random sample from the original distribution. Can you explain to me what it means? $\endgroup$
    – ccmatyn
    Nov 1, 2021 at 16:05
  • $\begingroup$ @ccmatyn Ah, that's another different application of weighted distributions. It occurs in sampling theory. If you want to represent an underlying population, but you have more observations than proportionally required of one part of it and less of another part, you have observed a weighted form of the original distribution with higher weights for parts that are more likely to be in your sample (and you may want to downweight these when estimating the underlying population distribution). $\endgroup$ Nov 1, 2021 at 16:11
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You can see this as a way of constructing new distributions from old ones. Disregarding your requirement that $X\ge 0$, let $X$ be a standard normal variable, $\phi(x)$ its density and $\Phi(x)$ its cdf. Then we have with $$w(x)=2\Phi(\alpha x)$$ that your construction becomes the skew-normal density.

You can replace the normal density above with other densities, and get other examples.

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