What is the expected value and the mean of sample standard deviation? I know that I can derive the expectation and variance of sample variance using the $\chi^2$ pdf. But I don't know how to start with sample standard deviation.
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According to the Wikipedia article on  the sample SD   $$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \overline{x})^2}$$  is a biased estimator of the SD of the population. It states that $E(\sqrt{s^2}) \neq \sqrt{E(s^2)}$.   NB. Random variables are independent and each $x_{i} \sim N(\mu,\sigma^{2})$   My question is two-fold:   What is the proof of the biasedness? How does one compute the expectation of the sample standard deviation    My knowledge of maths/stats is only intermediate.
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I have a distribution $X$. By playing around with random samples from $X$, I've determined that $Var(X^i) &gt; Var(X)$ where $i &gt; 1$. However, I can't seem to find a formula for the expected variance of $X^i$, or why it should be greater than $Var(X)$.  Moving away from normal distributions, should it generalize that the scale parameter of any $X^i$ will be greater than $X$?
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I am sampling from a parameter with unknown distribution. I would like to calculate a 95% CI for the standard deviation of the sample.   @cardinal provides a nice general solution for calculating a CI in his [answer] to my previous question,  . And @erik-p provides an estimate of the standard deviation of the variance of the sample.  However, in order to calculate the 95%CI for the sample variance, it seems that I need to know the distribution of the sample variance. Is it possible to calculate such an estimate without knowing the distribution from which the sample was taken?  A related question is
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I read from  that the standard error of the sample variance is  $$SE_{s^2} = \sqrt{\frac{2 \sigma^4}{N-1}}$$  What is the standard error of the sample standard deviation?  I'd be tempted to guess and say that $SE_{s} = \sqrt{SE_{s^2}}$ but I am not sure.
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I've been trying to compute a 95% confidence interval for the mean of a height sample, which is uniformly distributed. I have calculated the following sample statistics: $$n=10 \quad \quad \bar{x} = 73.55 \quad \quad s = 1.68 \quad \quad s^2 = 2.83.$$ How can I calculate the CI of the mean of a uniform distribution not knowing the limits of the distribution?  I only know the CI for the normal distribution. Thanks in advance.
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What is the precise difference between the Mean, Average and Expectation ? Is there any mathematical formaulation to show the difference ? I see that often they are used interchangeably.
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I got a problem of calculating $E[e^X]$, where X follows a normal distribution $N(\mu, \sigma^2)$ of mean $\mu$ and standard deviation $\sigma$.  I still got no clue how to solve it. Assume $Y=e^X$. Trying to calculate this value directly by substitution $f(x) = \frac{1}{\sqrt{2\pi\sigma^2}}\, e^{\frac{-(x-\mu)^2}{2\sigma^2}}$ then find $g(y)$ of $Y$ is a nightmare (and I don't know how to calculate this integral to be honest).  Another way is to find the inverse function. Assume $Y=\phi(X)$, if $\phi$ is differentiable, monotonic, and have inverse function $X=\psi(Y)$ then $g(y)$ (PDF of random variable $Y$) is as follows: $g(y)=f[\psi(y)]|\psi'(y)|$.  I think we don't need to find PDF of $Y$ explicitly to find $E[Y]$. This seems to be a classic problem. Anyone can help?
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I am trying to understand the relationship of a Random Variable to a population and random samples. I understand that a Random Variable is a function that maps each event in a sample space to a number. In this context, what would be a sample space and mapping? Would each sample be governed by the probability mass function?  Do we think of each sample as a Random Variable that takes on values from a population (sample space) randomly?   Or do we think of a population as a Random Variable and our sample as a random collection of those points?  I would appreciate if you can explain this to someone who is new to mathematical statistics. Thank you!
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Does someone know what the integral of the square of the normal distribution with standard deviation sigma over the range $-\infty$ to $+\infty$ is?  Thanks!
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[] ** calculate the mean and standard deviation of each investment, assume that both returns come from independent samples, are both normally distributed and that the (true but unknown) variances are the same,, Can someone could check my  calculations for  mean and standard deviation? I did calculations but not sure what to do with negative values of distances(If if am correct Investment A MEAN= 78.6875 when it comes to calculating standard deviation i have to power difference by 2 ?  **
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I'm learning applications on Central Limit Theorem and got really confused with a few points. According to this , the procedure to apply CLT usually goes like this:   So if SD is the population standard deviation, how are we gonna get it?? Isn't the whole population standard deviation what we eventually calculate by applying CLT and analyzing a sample of the whole population? How come the population standard deviation become a prerequisite??   Please tell me this tutorial is wrong.   I think the SD actually refers to the standard deviation of a sample (of some size n), which we can actually get easily. For example:    sample the whole population with a sample size of n (e.g. randomly select 10000 users from the whole population of 10 billions) calculate the mean of the 10000 measurements. calculate SE = SD / sqrt(10000), where SD is the standard deviation of the 10000 measurements, instead of the standard deviation of the whole population of 10 billions.    This explanation would make much more sense.   Any thoughts?
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Binomial problems: Mean and standard deviation  Suppose that the New England Colonials baseball team is equally likely to win any particular game as not to win it. Suppose also that we choose a random sample of Colonials games. Estimate the number of games in the sample that the Colonials win by giving the mean of the relevant distribution (that is, the expectation of the relevant random variable). Do not round your response. Quantify the uncertainty of your estimate by giving the standard deviation of the distribution. Round your response to at least three decimal places.  I've calculated the mean to be $10 (np=20*0.5)$. The short cut for calculating the variance of a binomial distribution is $np(1-p)$, but can you show me how to use the standard discrete distribution formula for variance ($Var(X) = E(X^2) - \mu^2$)to calculate variance in this case?
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In a problem that I'm solving I find that:  "Let data (yi,xi) be sampled randomly from a two-dimensional distribution such that y|x is N(ɑ,x^2σ^2)".  Are y and x i.i.d? maybe just identically distributed but not independent?  Finally, if my model is: y = ɑ - u (where u is the error term and ɑ is a constant), and I also know that the distribution of x does not depend on either a or σ (I don't know it this assumption is needed for answer at the following question), is u independent from x?  My question is due to the fact that I thought that if the data are sampled randomly this mean that the r.v. are independent, but how is possible that the conditional distribution (his variance) depend then on x?
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I was wondering what the difference between the variance and the standard deviation is.   If you calculate the two values, it is clear that you get the standard deviation out of the variance, but what does that mean in terms of the distribution you are observing?  Furthermore, why do you really need a standard deviation?
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Given a random variable $X\sim\mathcal N(0,\sigma^2)$, how can we prove that $E[X^4]=3\sigma^4$? I am having trouble even starting with the proof.
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I am trying to calculate the variance and standard deviation for a log normal distribution.  I was able to calculate the mean after reading this stack exchange article .  Now I want to calculate the variance and standard deviation.  I am unclear what formulas I can use.  I will calculate this in R.
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I am studying statistics and I am wondering when it comes to standard error or a sampling if the calculation changes when there are weights added.   I have a weighted mean:  $$\mu_{w} = \dfrac{\sum_{i=1}^Nw_ix_i}{\sum_{i=1}^Nw_i}$$ and a weighted variance calculated by:  $$s^2_{w}=\dfrac{\sum_{i=1}^Nw_i}{(\sum_{i=1}^Nw_i)^2-\sum_{i=1}^Nw_i^2}\cdot \sum_{i=1}^N(x_i-\mu)^2$$  is the sampling error still calculated as  $$\text{SE}=\sqrt{\dfrac{s^2_{w}}{n}}$$