Tag Info

New answers tagged

0

What you refer to as an incremental computation is very close to the computer scientist's notion of an online algorithm. There is in fact a well-known online algorithm for computing the variance and thus square root of a sequence of data, documented here.


0

Another way of saying the above is that you need to keep a count, an incremental sum of values and an incremental sum of squares. Let $N$ be the count of values seen so far, $S = \sum_{1,N} x_i$ and $Q = \sum_{1,N} x_i^2$ (where both $S$ and $Q$ are maintained incrementally). Then any stage, the mean is $\frac{S}{N}$ and the variance is $\frac{Q}{N} - ...


0

No, $\sigma$ is not the average $\ell^2$ distance from $\mu$. The $\ell^2$ distance from $x$ to $\mu$ is $\sqrt{(x-\mu)^2} = |x-\mu|$ and the average of those is $$ \sum_{x\in X} p(x)|x-\mu|. $$ Rather $\sigma$ is the $\ell^2$ distance from the tuple of $x$ values to the tuple in which every component is $\mu$. A reason for the use of the mean squared ...


1

Fitting a model such $$ y= \sum_{k=1}^{m} a_k\,x^{b_k}$$ using $n$ data points $(x_i,y_i)$ , $n \gt 2m$, is difficult because the model is nonlinear with respect to its parameters (the $b_k$'s) and nonlinear regression require reasonable starting guesses. As you noticed, the problem is simple if you assign specific values to the $b_k$'s since the problem ...


1

Note that the standard form of the Gaussian is $$pe^{-\dfrac{(x-q)^2}{2r^2}}$$ In your equation, we have $-\frac{1}{2r^2}=-b^2$, thus $r^2=\frac{1}{2b^2}$, thus $r=\sigma=\pm\sqrt{\frac{1}{2b^2}}$. However, the standard deviation is always positive. Therefore the results are: $$\sigma=\sqrt{\frac{1}{2b^2}}=\frac{\sqrt{2}}{2b}$$ $$\mathrm{FWHM} = 2 ...


0

Effects of sunshine, rain and humidity on the sales of various products correspond to some parameters of your statistical model, which you, then, each estimate by some function $f_i$ of the data in the sample. [of course, to be a good guess of the real value of parameter, the function $f_i$ has to be carefully chosen — that's what statistical theory is ...


1

Step 1 Use "$n-1$". See http://stats.stackexchange.com/questions/3931/intuitive-explanation-for-dividing-in-n-1-when-calculating-sd . If you use $n$ you have a biased estimator of the s.d. Since $n> n-1$ the bias is negative (towards too small s.d.s). I.e., use $\sigma_m \approx \sigma/\sqrt{n-1}$. Step 2 (Added by edit on 20150715, based on ...


1

The expectation is $$E[X]=K_1+K_2(t-10).$$ The variance is $$\sigma_X^2=E[(X-E[X])^2]=E[\{K_1+K_2(t-10)-(K_1+K_2(t-T_b)\}^2]=$$$$=K_2^2E[(10-T_b)^2]=25K_2^2.$$ So, the standard deviation $$\sigma_X=5K_2.$$


0

Note that for $x_i \not\in I$ we have $|x_i -x| \ge 3 \sigma$. Hence, \begin{align*} \sigma^2 &= \frac 1n \sum_{i=1}^n (x_i- x)^2\\ &\ge \frac 1n \sum_{x_i \not\in I} (x_i - x)^2\\ &\ge \frac 1n k \cdot 9\sigma^2\\ \iff n &\ge 9k \end{align*} The percentage of items in $I$, is by definition, $$ \frac {n-k}n \cdot ...


3

Test your sum of square $ s = 60.84 + \dots 51.84$. I get it as $s = 1003.20$. Then depending of the standard deviation type you get for the sample std. dev. $$\sigma_s = \sqrt{s/19} = 7.26636$$ or the population std. dev. $$\sigma_p = \sqrt{s/20} = 7.08237$$


4

There are two ways to calculate variance. First one is biased estimator and other is unbiased estimator of variance. In e-gadgets, probably estimator is unbiased estimator and that's what you are getting inconsistent result. Try dividing variance by $n-1$ , the number of data points instead of $n$ and then take square root. Good Luck!



Top 50 recent answers are included