Apa arti dari pooled variance "sebenarnya"?

15

Saya seorang noob dalam statistik, jadi bisakah kalian tolong bantu saya di sini.

Pertanyaan saya adalah sebagai berikut: Apa sebenarnya varian yang dikumpulkan? ?

Ketika saya mencari rumus untuk varian gabungan di internet, saya menemukan banyak literatur menggunakan rumus berikut (misalnya, di sini: http://math.tntech.edu/ISR/Mathematical_Statistics/Introduction_to_Statistics_Tests/thispage/newnode19.html ):

S_{p}^{2} = \frac{S_{1}^{2} (n_{1} - 1) + S_{2}^{2} (n_{2} - 1)}{n_{1} + n_{2} - 2}

$\begin{equation} \label{eq:stupidpooledvar} \displaystyle S^2_p = \frac{S_1^2 (n_1-1) + S_2^2 (n_2-1)}{n_1 + n_2 - 2} \end{equation}$

Tapi apa yang sebenarnya terjadi dihitung? Karena ketika saya menggunakan rumus ini untuk menghitung varian gabungan saya, itu memberi saya jawaban yang salah.

Misalnya, pertimbangkan "sampel induk" ini:

2, 2, 2, 2, 2, 8, 8, 8, 8, 8

$\begin{equation} \label{eq:parentsample} 2,2,2,2,2,8,8,8,8,8 \end{equation}$

Varian dari sampel induk ini adalah , dan rata-ratanya adalah . $S^2_p=10$ $\bar{x}_p=5$

Sekarang, misalkan saya membagi sampel induk ini menjadi dua sub-sampel:

Sub-sampel pertama adalah 2,2,2,2,2 dengan rata-rata dan varians $\bar{x}_1=2$ $S^2_1=0$ .
Sub-sampel kedua adalah 8,8,8,8,8 dengan rata-rata dan varians . $\bar{x}_2=8$ $S^2_2=0$

Sekarang, jelas, menggunakan rumus di atas untuk menghitung varian pooled / parent dari dua sub-sampel ini akan menghasilkan nol, karena dan . Jadi, apa rumus ini sebenarnya menghitung? $S_1=0$ $S_2=0$

Di sisi lain, setelah beberapa derivasi panjang, saya menemukan rumus yang menghasilkan varian pooled / parent yang benar adalah:

S_{p}^{2} = \frac{S_{1}^{2} (n_{1} - 1) + n_{1} d_{1}^{2} + S_{2}^{2} (n_{2} - 1) + n_{2} d_{2}^{2}}{n_{1} + n_{2} - 1}

$\begin{equation} \label{eq:smartpooledvar} \displaystyle S^2_p = \frac{S_1^2 (n_1-1) + n_1 d_1^2 + S_2^2 (n_2-1) + n_2 d_2^2} {n_1 + n_2 - 1} \end{equation}$

Dalam rumus di atas, dan $d_1=\bar{x_1}-\bar{x}_p$ $d_2=\bar{x_2}-\bar{x}_p$ .

Saya menemukan formula yang sama dengan milik saya, misalnya di sini: http://www.emathzone.com/tutorials/basic-statistics/combined-variance.html dan juga di Wikipedia. Meskipun saya harus mengakui bahwa mereka tidak persis sama dengan milik saya.

Jadi sekali lagi, apa arti sebenarnya dari kumpulan gabungan? Bukankah itu berarti varians sampel induk dari dua sub-sampel? Atau saya benar-benar salah di sini?

Terima kasih sebelumnya.

EDIT 1: Seseorang mengatakan bahwa dua sub-sampel saya di atas adalah patologis karena mereka memiliki nol varians. Baiklah, saya bisa memberi Anda contoh berbeda. Pertimbangkan contoh induk ini:

1, 2, 3, 4, 5, 46, 47, 48, 49, 50

$\begin{equation} \label{eq:parentsample2} 1,2,3,4,5,46,47,48,49,50 \end{equation}$

Varian dari sampel induk ini adalah , dan rata-ratanya adalah . $S^2_p=564.7$ $\bar{x}_p=25.5$

Sekarang, misalkan saya membagi sampel induk ini menjadi dua sub-sampel:

Sub-sampel pertama adalah 1,2,3,4,5 dengan rata-rata dan varian . $\bar{x}_1=3$ $S^2_1=2.5$
Sub-sampel kedua adalah 46,47,48,49,50 dengan rerata dan varians . $\bar{x}_2=48$ $S^2_2=2.5$

Sekarang, jika Anda menggunakan "rumus literatur" untuk menghitung varians yang dikumpulkan, Anda akan mendapatkan 2,5, yang sepenuhnya salah, karena varians induk / gabungan seharusnya 564,7. Sebaliknya, jika Anda menggunakan "formula saya", Anda akan mendapatkan jawaban yang benar.

Tolong mengerti, saya menggunakan contoh ekstrim di sini untuk menunjukkan kepada orang-orang bahwa formula itu memang salah. Jika saya menggunakan "data normal" yang tidak memiliki banyak variasi (kasus ekstrim), maka hasil dari kedua rumus tersebut akan sangat mirip, dan orang-orang dapat mengabaikan perbedaan karena kesalahan pembulatan, bukan karena rumus itu sendiri salah.

variance mean pooling

— Hanciong
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Beberapa tautan terkait untuk membantu: stats.stackexchange.com/q/214834/3277 , stats.stackexchange.com/q/12330/3277 , stats.stackexchange.com/q/43159/3277 .

— ttnphns

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Sederhananya, varians gabungan adalah perkiraan (tidak bias) dari varians dalam setiap sampel, di bawah asumsi / kendala bahwa varians tersebut sama.

Ini dijelaskan, dimotivasi, dan dianalisis secara rinci dalam entri Wikipedia untuk varian gabungan .

It does not estimate the variance of a new "meta-sample" formed by concatenating the two individual samples, like you supposed. As you have already discovered, estimating that requires a completely different formula.

— Jake Westfall
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The assumption of "equality" (that is, same population realized those samples) is not necessary in general to define what it is - "pooled". Pooled simply means averaged , omnibus (see my comment to Tim).

— ttnphns

@ttnphns I think the equality assumption is necessary for giving the pooled variance a conceptual meaning (which the OP asked for) that goes beyond just verbally describing the mathematical operation it performs on the sample variances. If the population variances are not assumed equal, then it's unclear what we could consider the pooled variance to be an estimate of. Of course, we could just think about it as being an amalgamation of the two variances and leave it at that, but that's hardly enlightening in the absence of any motivation for wanting to combine the variances in the first place.

— Jake Westfall

Jake, I'm not in disagreement with that, given the specific question of the OP, but I wanted to speak about definition of the word "pooled", that's why I said, "in general".

— ttnphns

@JakeWestfall Your answer is the best answer so far. Thank you. Although I am still not clear about one thing. According to Wikipedia, pooled variance is a method for estimating variance of several different populations when the mean of each population may be different, but one may assume that the variance of each population is the same.

— Hanciong

@JakeWestfall: Jadi jika kita menghitung varians yang dikumpulkan dari dua populasi yang berbeda dengan cara yang berbeda, apa yang sebenarnya dihitung? Karena varian pertama mengukur variasi sehubungan dengan mean pertama, dan varian kedua adalah sehubungan dengan mean kedua. Saya tidak tahu informasi tambahan apa yang dapat diperoleh dari menghitungnya.

— Hanciong

10

Varians gabungan digunakan untuk menggabungkan varian dari sampel yang berbeda dengan mengambil rata-rata tertimbang mereka, untuk mendapatkan varian "keseluruhan". Masalah dengan contoh Anda adalah bahwa itu adalah kasus patologis, karena masing-masing sub-sampel memiliki varians sama dengan nol. Kasus patologis seperti itu memiliki sangat sedikit kesamaan dengan data yang biasanya kita temui, karena selalu ada beberapa variabilitas dan jika tidak ada variabilitas, kami tidak peduli dengan variabel seperti itu karena mereka tidak membawa informasi. Anda perlu memperhatikan bahwa ini adalah metode yang sangat sederhana dan ada cara yang lebih rumit untuk memperkirakan varians dalam struktur data hierarkis yang tidak rentan terhadap masalah seperti itu.

As about your example in the edit, it shows that it is important to clearly state your assumptions before starting the analysis. Let's say that you have $n$ data points in $k$ groups, we would denote it as $x_{1,1},x_{2,1},\dots,x_{n-1,k},x_{n,k}$ , where the $i$ -th index in $x_{i,j}$ stands for cases and $j$ -th index stands for group indexes. There are several scenarios possible, you can assume that all the points come from the same distribution (for simplicity, let's assume normal distribution),

\begin{matrix} (1) & x_{i, j} \sim N (μ, σ^{2}) \end{matrix}

$x_{i,j} \sim \mathcal{N}(\mu, \sigma^2) \tag{1}$

you can assume that each of the sub-samples has its own mean

\begin{matrix} (2) & x_{i, j} \sim N (μ_{j}, σ^{2}) \end{matrix}

$x_{i,j} \sim \mathcal{N}(\mu_j, \sigma^2) \tag{2}$

or, its own variance

\begin{matrix} (3) & x_{i, j} \sim N (μ, σ_{j}^{2}) \end{matrix}

$x_{i,j} \sim \mathcal{N}(\mu, \sigma^2_j) \tag{3}$

or, each of them have their own, distinct parameters

\begin{matrix} (4) & x_{i, j} \sim N (μ_{j}, σ_{j}^{2}) \end{matrix}

$x_{i,j} \sim \mathcal{N}(\mu_j, \sigma^2_j) \tag{4}$

Depending on your assumptions, particular method may, or may not be adequate for analyzing the data.

In the first case, you wouldn't be interested in estimating the within-group variances, since you would assume that they all are the same. Nonetheless, if you aggregated the global variance from the group variances, you would get the same result as by using pooled variance since the definition of variance is

V a r (X) = \frac{1}{n - 1} \sum_{i} (x_{i} - μ)^{2}

$\mathrm{Var}(X) = \frac{1}{n-1} \sum_i (x_i - \mu)^2$

and in pooled estimator you first multiply it by $n-1$ , then add together, and finally divide by $n_1 + n_2 - 1$ .

In the second case, means differ, but you have a common variance. This example is closest to your example in the edit. In this scenario, the pooled variance would correctly estimate the global variance, while if estimated variance on the whole dataset, you would obtain incorrect results, since you were not accounting for the fact that the groups have different means.

In the third case it doesn't make sense to estimate the "global" variance since you assume that each of the groups have its own variance. You may be still interested in obtaining the estimate for the whole population, but in such case both (a) calculating the individual variances per group, and (b) calculating the global variance from the whole dataset, can give you misleading results. If you are dealing with this kind of data, you should think of using more complicated model that accounts for the hierarchical nature of the data.

The fourth case is the most extreme and quite similar to the previous one. In this scenario, if you wanted to estimate the global mean and variance, you would need a different model and different set of assumptions. In such case, you would assume that your data is of hierarchical structure, and besides the within-group means and variances, there is a higher-level common variance, for example assuming the following model

\begin{matrix} (5) & \begin{aligned} x_{i, j} & \sim N (μ_{j}, σ_{j}^{2}) \\ μ_{j} & \sim N (μ_{0}, σ_{0}^{2}) \\ σ_{j}^{2} & \sim I G (α, β) \end{aligned} \end{matrix}

$\begin{align} x_{i,j} &\sim \mathcal{N}(\mu_j, \sigma^2_j) \\ \mu_j &\sim \mathcal{N}(\mu_0, \sigma^2_0) \\ \sigma^2_j &\sim \mathcal{IG}(\alpha, \beta) \end{align} \tag{5}$

where each sample has its own means and variances $\mu_j,\sigma^2_j$ that are themselves draws from common distributions. In such case, you would use a hierarchical model that takes into consideration both the lower-level and upper-level variability. To read more about this kind of models, you can check the Bayesian Data Analysis book by Gelman et al. and their eight schools example. This is however much more complicated model then the simple pooled variance estimator.

— Tim
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I have updated my question with different example. In this case, the answer from "literature's formula" is still wrong. I understand that we are usually dealing with "normal data" where there is no extreme case like my example above. However, as mathematicians, shouldn't you care about which formula is indeed correct, instead of which formula applies in "everyday/common problem"? If some formula is fundamentally wrong, it should be discarded, especially if there is another formula which holds in all cases, pathological or not.

— Hanciong

Btw you said there are more complicated ways of estimating variance. Could you show me these ways? Thank you

— Hanciong

2

Tim, pooled variance is not the total variance of the "combined sample". In statistics, "pooled" means weighted averaged (when we speak of averaged quantities such as variances, weights being the n's) or just summed (when we speak of sums such as scatters, sums-of-squares). Please, reconsider your terminology (choice of words) in the answer.

— ttnphns

1

Albeit off the current topic, here is an interesting question about "common" variance concept. stats.stackexchange.com/q/208175/3277

— ttnphns

1

Hanciong. I insist that "pooled" in general and even specifically "pooled variance" concept does not need, in general, any assumption such as: groups came from populations with equal variances. Pooling is simply blending (weighted averaging or summing). It is in ANOVA and similar circumstances that we do add that statistical assumption.

— ttnphns

1

The problem is if you just concatenate the samples and estimate its variance you're assuming they're from the same distribution therefore have the same mean. But we are in general interested in several samples with different mean. Does this make sense?

— ZHU
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0

The use-case of pooled variance is when you have two samples from distributions that:

may have different means, but
which you expect to have an equal true variance.

An example of this is a situation where you measure the length of Alice's nose $n$ times for one sample, and measure the length of Bob's nose $m$ times for the second. These are likely to produce a bunch of different measurements on the scale of millimeters, because of measurement error. But you expect the variance in measurement error to be the same no matter which nose you measure.

In this case, taking the pooled variance would give you a better estimate of the variance in measurement error than taking the variance of one sample alone.

— Misha
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Thank you for your answer, but I still don't understand about one thing. The first data gives you the variance with respect to Alice's nose length, and the second data gives you the variance with respect to Bob's nose length. If you are calculating a pooled variance from those data, what does it mean actually? Because the first variance is measuring the variation with respect to Alice's, and the second with respect to Bob's, so what additional information can we gained by calculating their pooled variance? They are completely different numbers.

— Hanciong

0

Through pooled variance we are not trying to estimate the variance of a bigger sample, using smaller samples. Hence, the two examples you gave don't exactly refer to the question.

Pooled variance is required to get a better estimate of population variance, from two samples that have been randomly taken from that population and come up with different variance estimates.

Example, you are trying to gauge variance in the smoking habits of males in London. You sample two times, 300 males from London. You end up getting two variances (probably a bit different!). Now since, you did a fair random sampling (best to your capability! as true random sampling is almost impossible), you have all the rights to say that both the variances are true point estimates of population variance (London males in this case).

But how is that possible? i.e. two different point estimates!! Thus, we go ahead and find a common point estimate which is pooled variance. It is nothing but weighted average of two point estimates, where the weights are the degree of freedom associated with each sample.

Hope this clarifies.

— Sameer Saurabh
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