दो या दो से अधिक समूहों के पूल किए गए विचरण की गणना ज्ञात समूह संस्करण, साधन और नमूना आकार कैसे करें?

मान लें कि $m+n$ तत्व दो समूहों ( $m$ और $n$ ) में विभाजित हैं । पहले समूह का विचरण है $\sigma_m^2$ और दूसरे समूह के विचरण है $\sigma^2_n$ । तत्वों को अज्ञात माना जाता है, लेकिन मुझे पता है कि $\mu_m$ और $\mu_n$ ।

वहाँ संयुक्त विचरण गणना करने के लिए एक रास्ता है $\sigma^2_{(m+n)}$ ?

विचरण के लिए निष्पक्ष होना आवश्यक नहीं है, इसलिए हर $(m+n)$ और नहीं $(m+n-1)$ ।

variance pooling

— user1809989
स्रोत

जब आप कहते हैं कि आप इन समूहों के साधन और प्रकार जानते हैं, तो क्या वे पैरामीटर या नमूना मूल्य हैं? वे नमूना साधन हैं / प्रसरण आप उपयोग नहीं करना चाहिए

और

...

μ

$\mu$

σ

$\sigma$

— जोनाथन क्रिस्टेंसेन

मैंने सिर्फ प्रतीकों को एक प्रतिनिधित्व के रूप में इस्तेमाल किया। अन्यथा, मेरी समस्या को समझाना कठिन होता।

— user1809989

नमूना मूल्यों के लिए, हम आमतौर पर लैटिन अक्षरों (जैसे

और

) का उपयोग करते हैं। ग्रीक अक्षर आमतौर पर मापदंडों के लिए आरक्षित होते हैं। "सही" (अपेक्षित) प्रतीकों का उपयोग करने से आपको अधिक स्पष्ट रूप से संवाद करने में मदद मिलेगी।

m

$m$

s

$s$

— जोनाथन क्रिस्टेंसन

कोई चिंता नहीं, मैं अभी से उस का पालन करूँगा! चीयर्स

— user1809989

@Jonathan क्योंकि इस नमूने या अनुमान के बारे में सवाल नहीं है, एक वैध तरीके से देखने का समय लग सकता है कि

और

हैं सच मतलब और डेटा का एक बैच के अनुभवजन्य वितरण के विचरण, जिससे यूनानी पत्र के पारंपरिक उपयोग को न्यायोचित ठहरा के बजाय उन्हें संदर्भित करने के लिए लैटिन पत्र।

μ

$\mu$

σ^{2}

$\sigma^2$

— whuber

जवाबों:

माध्य की परिभाषाओं का उपयोग करें

μ_{1 : n} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}

$\mu_{1:n} = \frac{1}{n}\sum_{i=1}^n x_i$

और नमूना विचरण

σ_{1 : n}^{2} = \frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - μ_{1 : n})}^{2} = \frac{n - 1}{n} (\frac{1}{n - 1} \sum_{i = 1}^{n} {(x_{i} - μ_{1 : n})}^{2})

$\sigma_{1:n}^2 = \frac{1}{n}\sum_{i=1}^n \left(x_i - \mu_{1:n}\right)^2 = \frac{n-1}{n}\left(\frac{1}{n-1}\sum_{i=1}^n \left(x_i - \mu_{1:n}\right)^2\right)$

(the last term in parentheses is the unbiased variance estimator often computed by default in statistical software) to find the sum of squares of all the data $x_i$ . Let's order the indexes $i$ so that $i=1,\ldots,n$ designates elements of the first group and $i=n+1,\ldots,n+m$ designates elements of the second group. Break that sum of squares by group and re-express the two pieces in terms of the variances and means of the subsets of the data:

\begin{aligned} (m + n) (σ_{1 : m + n}^{2} + μ_{1 : m + n}^{2}) & = \sum_{i = 1}^{1 : n + m} x_{i}^{2} \\ = \sum_{i = 1}^{n} x_{i}^{2} + \sum_{i = n + 1}^{n + m} x_{i}^{2} \\ = n (σ_{1 : n}^{2} + μ_{1 : n}^{2}) + m (σ_{1 + n : m + n}^{2} + μ_{1 + n : m + n}^{2}) . \end{aligned}

$\eqalign{ (m+n)(\sigma^2_{1:m+n} + \mu_{1:m+n}^2) &= \sum_{i=1}^{1:n+m} x_i^2 \\ &= \sum_{i=1}^n x_i^2 + \sum_{i=n+1}^{n+m} x_i^2 \\ &= n(\sigma^2_{1:n} + \mu_{1:n}^2) + m(\sigma^2_{1+n:m+n} + \mu_{1+n:m+n}^2). }$

Algebraically solving this for $\sigma^2_{m+n}$ in terms of the other (known) quantities yields

σ_{1 : m + n}^{2} = \frac{n (σ_{1 : n}^{2} + μ_{1 : n}^{2}) + m (σ_{1 + n : m + n}^{2} + μ_{1 + n : m + n}^{2})}{m + n} - μ_{1 : m + n}^{2} .

$\sigma^2_{1:m+n} = \frac{n(\sigma^2_{1:n} + \mu_{1:n}^2) + m(\sigma^2_{1+n:m+n} + \mu_{1+n:m+n}^2)}{m+n} - \mu^2_{1:m+n}.$

Of course, using the same approach, $\mu_{1:m+n} = (n\mu_{1:n} + m\mu_{1+n:m+n})/(m+n)$ can be expressed in terms of the group means, too.

An anonymous contributor points out that when the sample means are equal (so that $\mu_{1:n}=\mu_{1+n:m+n}=\mu_{1:m+n}$ ), the solution for $\sigma^2_{m+n}$ is a weighted mean of the group sample variances.

— whuber
स्रोत

The "homework" tag doesn't mean the question is elementary or stupid: it's used for self-study questions that can even include research-level queries. It distinguishes routine, more or less context-free questions (of the sort that might ordinarily grace the math forum) from specific applied questions.

— whuber

I cannot understand your first passage:

n (σ^{2} + μ^{2}) = \sum (x - μ)^{2} + n μ^{2} \overset{?}{=} \sum x^{2}

$n(\sigma^2+\mu^2) = \sum (x - \mu)^2 + n\mu^2 \stackrel{?}{=} \sum x^2$ In particular I get

\sum [(x - μ)^{2} + μ^{2}] = \sum [x^{2} - 2 x μ]

$\sum [(x-\mu)^2+\mu^2] = \sum [x^2-2x\mu]$ which requires

μ = 0

$\mu = 0$ Am I missing something? Could you please explain this?

— DarioP

@Dario

\sum (x - μ)^{2} + n μ^{2} = (\sum x^{2} - 2 μ \sum x + n μ^{2}) + n μ^{2} = \sum x^{2} - 2 n μ^{2} + 2 n μ^{2} = \sum x^{2} .

$\sum(x-\mu)^2+n\mu^2=(\sum x^2 - 2\mu\sum x + n \mu^2)+n\mu^2 = \sum x^2 - 2n\mu^2 + 2n\mu^2 = \sum x^2.$

— whuber

Oh yes, I did a stupid sign mistake in my derivation, now is clear, thanks!!

— DarioP

I guess this can be extended to an arbitrary number of samples as long as you have the mean and variance for each. Calculating pooled (biased) standard deviation in R is simply sqrt(weighted.mean(u^2 + rho^2, n) - weighted.mean(u, n)^2) where n, u and rho are equal-length vectors. E.g. n=c(10, 14, 9) for three samples.

— Jonas Lindeløv

I'm going to use standard notation for sample means and sample variances in this answer, rather than the notation used in the question. Using standard notation, another formula for the pooled sample variance of two groups can be found in O'Neill (2014) (Result 1):

\begin{aligned} s_{pooled}^{2} & = \frac{1}{n_{1} + n_{2} - 1} [(n_{1} - 1) s_{1}^{2} + (n_{2} - 1) s_{2}^{2} + \frac{n_{1} n_{2}}{n_{1} + n_{2}} ({\bar{x}}_{1} - {\bar{x}}_{2})^{2}] . \end{aligned}

$\begin{equation} \begin{aligned} s_\text{pooled}^2 &= \frac{1}{n_1+n_2-1} \Bigg[ (n_1-1) s_1^2 + (n_2-1) s_2^2 + \frac{n_1 n_2}{n_1+n_2} (\bar{x}_1 - \bar{x}_2)^2 \Bigg]. \\[10pt] \end{aligned} \end{equation}$

This formula works directly with the underlying sample means and sample variances of the two subgroups, and does not require intermediate calculation of the pooled sample mean. (Proof of result in linked paper.)

— Reinstate Monica
स्रोत

-3

Yes, given the mean, sample count, and variance or standard deviation of each of two or more groups of samples, you can exactly calculate the variance or standard deviation of the combined group.

This web page describes how to do it, and why it works; it also includes source code in Perl: http://www.burtonsys.com/climate/composite_standard_deviations.html

BTW, contrary to the answer given above,

\begin{aligned} n (σ^{2} + μ^{2}) \neq \sum_{i = 1}^{n} x_{i}^{2} \end{aligned}

$\eqalign{ n(\sigma^2 + \mu^2) \space\space \ne \space\space \sum_{i=1}^n x_i^2 }$

See for yourself, e.g., in R:

> x = rnorm(10,5,2)
> x
 [1] 6.515139 8.273285 2.879483 3.624233 6.199610 3.683164 4.921028 8.084591
 [9] 2.974520 6.049962
> mean(x)
[1] 5.320502
> sd(x)
[1] 2.007519
> sum(x**2)
[1] 319.3486
> 10 * (mean(x)**2 + sd(x)**2)
[1] 323.3787

— Dave Burton
स्रोत

it's because you forgot the n-1 factor, e.g. try with n*(mean(x)**2+sd(x)**2/(n)*(n-1))

— user603

user603, what on earth are you talking about?

— Dave Burton

Dave, mathematics is a more reliable teacher than software. In this case R computes the unbiased estimate of the standard deviation rather than the standard deviation of the set of numbers. For instance, sd(c(-1,1)) returns 1.414214 rather than 1. Your example needs to use sqrt(9/10)*sd(x) in place of sd(x). Interpreting "

σ

$\sigma$ " as the SD of the data and "

μ

$\mu$ " as the mean of the data, your BTW remark is wrong. A program demonstrating this is n <- 10; x <- rnorm(n,5,2); m <- mean(x); s <- sd(x) * sqrt((n-1)/n); m2 <- sum(x^2); c(lhs=n * (m^2 + s^2), rhs=m2)

— whuber