# Tag Info

0

That is the non-probabilistic (statistical) version of the Chebyshev inequality. Let $S$ be the "tail set": $i\in S \iff |x_i - \overline{x}|\ge a$, and let $|S|=k$. Then $$\sum_{i=1}^n(x_i - \overline{x})^2= \sum_{i\in S} (x_i - \overline{x})^2+\sum_{i\notin S} (x_i - \overline{x})^2=\\ \ge \sum_{i\notin S} (x_i - \overline{x})^2 \ge k \,a^2$$ Then, ...

1

The Standard Deviation is a measure of how spread out the numbers in your data are. So if your numbers have the same difference, even with higher values then you will got the same standard daviation. A small example to clarify, using matlab: If your data is : $1 ,2,3,4,5,6,7,8,9$ then std is $2.7386$. If your data is : $10 ,20,30,40,50,60,70,80,90$ ...

0

The variance, which is the square of the standard deviation, is $\sigma^2=\overline{S^2}-(\overline{S})^2$. The expected value of $S$ is $\overline S=\sum_i\frac NkE(X_i)=\sum_i\frac Nk\frac MN$. The expected value of $S^2$ is $$\overline{S^2}=\frac{N^2}{k^2}\left(kE(X_i^2)+k(k-1)E_{i\neq j}(X_iX_j)\right)$$ That is because there are $k^2$ terms in $S^2$, ...

0

The way this usually goes is slightly different from what you've written. Usually we assume we have a bunch of independent Gaussian variables $X_1,\dots,X_n$ all of which have mean $\mu$ and which have (perhaps unknown) standard deviations $\sigma_i$. If we know $\sigma_i$ then the least squares estimator of $\mu$ is $$\sum_{k=1}^n \frac{X_i}{\sigma_i^2}.$$ ...

0

HINT So let $m,s$ be mean and std dev, you have $k=3$ and your equations yield $$m + 3s = 438572\\ m - 3s = 189992$$ Can you solve 2 equations for 2 unknowns? HINT 2 What happens when you add the equations to each other?

2

I decided to move from a comment to an answer because some times I ended up using some facts without digging too much in the whys. So, I decided to dig a little more and hopefully answer the questions of the OP. This answer is mainly about why the RMS of the noise is equal to its standard deviation. As a side note, I will also do some comments about the ...

0

The method to find the percentage $p$ of values above a certain point $a$ in a normally distributed set with the mean $\mu$ and standard deviation $\sigma$ is to integrate the normal distribution from $a$ to $\infty$. Thus this gives $$p=\int_a^\infty \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$ To find the percentage of values below $a$, ...

7

Let's replace $Var(X)$ with $\sigma^2$ in the first equation to give $$P(|X - E(X)| \geq r) \leq \frac{\sigma^2}{r^2}.$$ Now suppose $k= \dfrac{r}{\sigma}$, i.e. $r = k \sigma$, and substitute to give $$P(|X - E(X)| \geq k \sigma) \leq \frac{\sigma^2}{k^2\sigma^2} = \frac{1}{k^2}$$ which is your second equation using $k$ instead of $r$. You can think of ...

1

Note that $Var(X) = \sigma^2$. \begin{align*} P(|X - E(X)| \geq r) \leq \frac{\sigma^2}{r^2} &\implies P((X - E(X))^2 \geq r^2) \leq \frac{\sigma^2}{r^2} \\ &\implies P((X - E(X))^2 \geq r^2\sigma^2) \leq \frac{1}{r^2} \\ &\implies P(|X - E(X)| \geq r\sigma) \leq \frac{1}{r^2} \end{align*}

3

It is exactly the same. If you use the first inequality you have $$P(|X−E(X)|\geq r⋅σ)\leq \frac{\operatorname{var}(X)}{(r\sigma)^2}=\frac{1}{r^2}.$$

0

Variance($V$) of $(X-Y)=V(X)+V(Y)-2Cov(X,Y)$ where $Cov(X,Y)$ is the covariance between $X$ and $Y$. Mean($\mu$) of $(X-Y)=\mu(X)-\mu(Y)$ So you need information regarding the covariance to calculate the answers properly if $X$ and $Y$ are not independent. EDIT: $$Cov(X,Y)=\sum_{i=1}^{n}\frac{(x_i-\bar x)(y_i-\bar y)}{\sigma_x\sigma_y}$$

0

A 3-d multivariate distribution would be the same as the 3 independent 1-d distributions if, you guessed, the dimensions are statistically independent of each other. If they are correlated, then you have to introduce that correlation one way or another.

0

Sample variance also i depends on population size: when you write out the formula for $\sigma^2/n$ you can see the expression for $\sigma$ will have population size. Using $N$ for population size, $\mu$ for population mean: Sample variance is $\frac{1}{n\sqrt N}\sqrt {\sum_{j=1}^N(x_j-\mu)^2}$

1

This isn't really an intuitive approach. Let $X$ be a random variable such that $X \sim N(\mu,\sigma^2)$ If we have a sample of $n$ independent observations of $X$ then $$X_1 +X_2 +X_3+ \dots +X_n \sim N(n\mu,n\sigma^2)$$ $$\text{sample mean} = \bar X = \frac{X_1 +X_2 +X_3+ \dots +X_n}{n} \sim N(\mu,\frac{\sigma^2}{n})$$ Note that $aX \sim ... 2 Neither of these is the actual definition of standard deviation. If you have a random variable$X$, the standard deviation$\sigma$is defined by $$\sigma^2 = E [ (X-EX)^2 ]$$ where$E$denotes the expected value (i.e., if$X$takes$k$values$a_1, ..., a_k$with probabilities$p_1,...,p_k$respectively,$EX=\sum_{i=1}^k p_i a_i$). Of course,$\sigma^2\$ ...

1

The first one is used for population and second one is for sample. What is the difference between a population and a sample? A sample is a subset of people, items, or events from a larger population that you collect and analyze to make inferences. To represent the population well, a sample should be randomly collected and adequately large. To understand ...

Top 50 recent answers are included