Linked Questions

2
votes
2answers
40 views

“… because sample mean gets different values from sample to sample and it is a random variable with mean $\mu$ and variance $\frac{\sigma^2}{n}$.”

This answer by user "sevenkul" says the following: The sample mean $\overline{X}$ also deviates from $\mu$ with variance $\frac{\sigma^2}{n}$ because sample mean gets different values from ...
1
vote
0answers
21 views

Different sample covariance formulae (conventions) [duplicate]

Page 358 of Introduction to probability, second edition, by Blitzstein and Hwang, defines the sample covariance as $$r = \dfrac{1}{n} \sum_{i = 1}^n (x_i - \bar{x})(y_i - \bar{y}),$$ where $\bar{x} = \...
1
vote
1answer
1k views

Covariance Matrix vs. Pairwise Covariance Matrix?

I found this equation here to calculate a covariance matrix of any number of variables using matrix algebra. $$\frac1{N} (X - 1\bar{x})^T(X - 1\bar{x}^T) $$ For a given matrix $X$ with $N$ samples. ...
0
votes
0answers
15 views

Degree of freedom [duplicate]

Why do we use degree of freedom while dealing with samples(e.g. while calculating t statistic and sample variance, we divide by $ n-1 $ degree of freedom, but while calculating z score and population ...
1
vote
0answers
19 views

Is there ever a right time to not use Bessel's Correction? [duplicate]

I can not find a clear explanation for when NOT to use Bessel's correction and use N instead of N-1. As I understand it, Bessel's correction would apply to things such as clinical trials, sampling ...
1
vote
0answers
28 views

Explanation for Bessel's correction [duplicate]

In textbooks and online the explanation for Bessel's correction is that the sample points are closer to the sample mean than the population mean.This is true for sampling with replacement as well. So, ...
0
votes
0answers
17 views

Show estimator is unbiased [duplicate]

Consider the estimator of the variance given by the formula: $(S')^2 = \frac{1}{n} \sum_{i=1}^{n}(Y_i − µ)^2$ Is this a biased or unbiased estimator? I'm not sure if it is possible to prove ...
1
vote
1answer
635 views

The derivation of standard deviation [duplicate]

So, if there are $N$ data points in my sample space of any randomly distributed variable $X$. The standard deviation, $\sigma$, (from my understanding) is the root mean squared of the error (from the ...
4
votes
2answers
214 views

Sample Variance and Dividing by $n-1$

In this video... https://www.youtube.com/watch?v=sHRBg6BhKjI ...and in many others, the explanation for why when calculating the sample variance we divide by $n-1$ instead of by $n$ is the following:...
0
votes
1answer
80 views

Why Standard deviation formula has n-1 in the denominator but Variance only n? [duplicate]

If is true that SQRT(Variance) = SD then one cannot have n-1 in the denominator and the other not but should be same? SD formula: Variance formula:
2
votes
1answer
137 views

Why is $\frac{\sum^n_{i=1}(X_i-\bar{X})^2}{\sigma^2}$chi-square distributed with $n-1$ degrees of freedom?

On Wikipedia, it says If $X_{i};i=1,\ldots ,n$ are independent normal $(\mu ,\sigma ^{2})$ random variables, the statistic $$\frac{\sum\limits^n_{i=1}(X_i-\bar{X})^2}{\sigma^2}$$ ...
0
votes
0answers
123 views

What are biased and inefficient estimators?

I’m studying statistics from Schaum’s Outline, which gives the following: ...
8
votes
3answers
2k views

When estimating variance, why do unbiased estimators divide by n-1 yet maximum likelihood estimates divide by n?

I am totally confused: On the one hand you can read all kinds of explanations why you have to divide by n-1 to get an unbiased estimator for the (unknown) population variance (degrees of freedom, not ...
2
votes
1answer
92 views

Why do I need to rounding x and y values to the nearest 0.5 when manually calculating correlation?

I'm trying to calculate correlation using a formula in Statistics 4th Edition by Freedman: r = average of (x in standard units) * (y in standard units) If I try this out ... ...
0
votes
1answer
223 views

Estimates of variance from an iid sample [duplicate]

There are two kinds of estimates of variance from an iid sample $X1, \dots, X_n$ $1/n * \sum_i (X_i - \bar{X})^2$, which is MLE $1/(n-1) * \sum_i (X_i - \bar{X})^2$, which is unbiased. The unbiased ...

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