New answers tagged variance
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The join-count statistic is a special case of a class that includes the Getis-Ord $G$, Moran, and Geary statistics [Haining, Spatial Data Analysis Theory and Practice (2003), p. 242]. They are united by a graph-theoretic abstraction in which spatial features are nodes, features $i$ and $j$ (with $i \ne j$) considered adjacent are joined by edges $e_{ij}$ ...
4
I like to try to "let the data do the talking" instead of "coming to it and telling it what to say".
I encourage you to perform Exploratory Data Analysis before culling any data.
If this were my data then I would consider making a bubble plot where the mean (or median) lake size is along the x-axis, the variation in lake size is along the y-axis, and the ...
2
There appears to be a difference in the interpretation of a statistical formula. One quick, simple, and compelling way to resolve such differences is to simulate the situation. Here, you have noted there will be a difference when the players play different numbers of games. Let's therefore retain every aspect of the question but change the number of games ...
0
One option is to report the ratio of the larger to the smaller standard deviations or the ratio of the larger to the smaller variance for the two groups. Such a metric is more comparable than the raw difference in standard deviation or variance with different studies that are on different scales.
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One of the properties of variance is
\begin{equation}
Var(AZ) = A^{2} Var(Z)
\end{equation}
where A is a constant and Z is a random variable.
The matrix equivalent is the version you're working with, i.e.
\begin{equation}
Var(AZ) = A Var(Z) A^{T}
\end{equation}
Two links for you to check out are Variance#properties on Wikipedia and also these course notes ...
0
I wouldn't place much stock in "rules of thumb" such as this. It is dependent upon so many things such as the number of variables, the number of sites, what dissimilarity you use etc. Also note that the vector fitting approach is inherently linear and we have no reason to presume that the relationship between the variable and the NMDS configuration is ...
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I think the key phrase to use when explaining both variance and standard deviation is "measure of spread". In the most basic language, the variance and standard deviation tell us how well spread out the data is. To be a little more accurate, although still addressing the layman, they tell us how well the data is spread out around the mean. In passing, note ...
-1
I'm sorry, but that book is using the terms incorrectly. $S^2$ is the sample variance, NOT population variance. $\sigma^2$ is the population variance. When estimating $\sigma^2$, you should always use $N$ as denominator, and never $N-1$. For sample variance $n-1$ should be always used because the sample variance underestimates the population variance.
...
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It could be possible. If you code an east-west variable, a simple binary variable. Check it's correlation with your mode variable. If they are very highly correlated, then multicollinearity may be at play. i.e. your mode variable may in fact be explaining away the east west divide.
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This stems from your basic regression model.
Two things.
1:
In the context of regression we assume we have a perfect, complete model ready to go which explains, correctly, all covariability in the observed data for all variables.
Implicit with this is that the model is assumed constant for all observed data. Intuitively we tend to think of a time series but ...
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I don't think equality of variance is natural in all circumstances; indeed we know many situations where it isn't the case.
However, in situations where a shift in the mean doesn't otherwise affect the distribution of the values, then you would expect to see it.
The assumption is sometimes one of convenience - the distribution of test statistics under the ...
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If you're talking about statistical tests of variance between two or more populations: I don't think we do, but I think we simply state such in the null hypothesis so we can calibrate a statistical test for the alternative, as is the common case in NHST. If indeed, we claim that $\mathcal{H}_0: \sigma^2_1 = \sigma^2_2$ then we can go on to devise a family of ...
2
Within-cluster-variance is a simple to understand measure of compactness (there are others, too).
So basically, the objective is to find the most compact partitioning of the data set into $k$ partitions.
K-Means, in the Lloyd version, actually originated from 1d PCM data as far as I know. So assuming you have a really bad telephone line, and someone is ...
1
There are several questions here at very different levels. In essence every text on cluster analysis is an answer. You have to keep reading!
Variance is at one level just one statistical standard which statistical people find convenient to think about. Roughly, minimising variance encourages -- nay, enforces -- clusters as relatively tight balls. What can ...
1
@Glen_b examples showed that a strong correlation is not strictly impossible in this situation but I think your intuition is right. Limited variation (for example range restriction) can bias sample correlations toward 0. The problem is not merely one of “significance”, the correlation also appears systematically smaller than it would be if you would consider ...
1
It's possible to get high point biserial correlation even with 27 $1$'s and a $0$. Indeed, you can get as high as 1, so it's not that:
y <- c(0,rep(1,27))
x <- y
cor(x,y)
[1] 1
-- and making x continuously distributed doesn't substantively alter that conclusion:
y <- c(0,rep(1,27))
x=c(rnorm(1,0),rnorm(27,100))
cor(x,y)
[1] 0.9987537
...
0
I found it at: Computational Finance of George Levy page appendix I.
1
Think about how many degrees of freedom there must be (total, residual & regression). Then you can work backwards from the $MSE$ to get the missing $SS$s.
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I am not an expert but I can say:
autoarima() will attempt to fit a integrated auto regressive moving average process to your data. Once the data is appropriately differenced and the auto regressive component and the moving average components are removed you are left with residuals with a variance. In R this is the sigma2 in the arima object. Usually I ...
3
There are two questions here.
The answer to your general question is "no", model error is not the same as bias. Bias is the difference between the expected value of your estimator and the true parameter.
In many instances of model error, the parameter will not even feature in the model. With a bit of effort you could reparameterise your model so that ...
0
Regarded as a question in probability theory, the answer to this question is that $$E[Y]=E[X_1+X_2+\cdots +X_n] = E[X_1] + E[X_2]+\cdots + E[X_n]$$
via a result known as the linearity of expectation, and since the random variables
all have zero mean in this particular instance, $E[Y]=0$. On the other hand,
the
information given is insufficient to ...
0
If the trials are independent, you're looking for the waiting time $T$ of a Bernoulli process with probability $p=0.85$ and failure $q=1-p$, which is a Geometric distribution. You can easily compute the mean and variance by these formulae,
$$\mathbb{E}(T) = \frac{1}{p}~,$$
$$\mathbb{V}(T) = \frac{q}{p^2}~.$$
0
Working off Peter Flom's suggestion, look at this R code:
x <- as.data.frame(matrix(rnorm(50000,0,1), nr = 5000, nc = 10))
x$y <- x$V1 + x$V2 + x$V3 + x$V4 + x$V5 + x$V6 + x$V7 + x$V8 + x$V9 + x$V10
We generate 10 variables (V1 through V10), each with a mean of 0 and a standard deviation of 1. Y is constructed to be the sum of the ten variables for ...
1
There is a great deal of disagreement over good statistical style here, and indeed most of elsewhere.
But this strikes me as a mishmash of quite different procedures.
No tests for differing variances will work as designed if you Winsorize the data first. Perhaps someone has worked on this -- you might find literature references with modified tests -- but ...
1
We frequently use scaling relationships between central moments to ascertain the most likely underlying generative process of biological data. I studied the dynamics of the HIV promoter and it produces very non-Gaussian distributions that are wide and highly skewed. The Var~Mean relationship was the first step in determining candidate stochastic models. ...
3
What do you mean 'still'? The covariance matrix is
$$
\boldsymbol{\Sigma}_u=
\begin{bmatrix}
E(u_1^2)&E(u_1u_2) & \cdots & E(u_1u_n)\\
E(u_1u_2)&E(u_2^2) & \cdots & E(u_2u_n)\\
\vdots & \vdots & \ddots & \vdots\\
E(u_1u_n)&E(u_2u_n) & \cdots & E(u_n^2)\\
\end{bmatrix}
$$
so under the assumptions that
$$
...
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