I was asked if I sample from a distribution with a standard deviation of 1, with a sample size of 5, what do I expect the standard deviation of my sample to be. I thought that it would be something weird, because I've head this talk of sample standard deviation being a biased estimator, but the solution they gave was that the sample standard deviation should be approximately the distribution's standard deviation.
@Anthony I think one can prove that in the limit that the sample size goes to infinity, the std dev of the sample equals that of the underlying distribution
I think they were looking for a very qualitative answer
@KevinDriscoll I thought the expectation of the sample standard deviation for fixed $n$ is less than the distribution deviation, this is the whole notion of it being a biased estimator isn't it?
According to the Wikipedia article on unbiased estimation of standard deviation 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 indepe...
@Anthony Yea that's right, sorry. I got confused about whether we're talking about the actual standard deviation of the sample form the mean of the sample, or what is sometimes called the sample standard deviation, which is actually the corrected estimate of the population standard deviation
Suppose I have two random vectors $\mathbf{Y}_1 \sim \mathcal{N}(\mathbf{\mu}_1, \boldsymbol{\Sigma}_1)$ and $\mathbf{Y}_2 \sim \mathcal{N}(\mathbf{\mu}_2, \boldsymbol{\Sigma}_2)$. If $Y_1$ and $Y_2$ are uncorrelated, I believe they are independent (but this only applies to normal distributions, I think). @MikeMiller
@MikeMiller FYI, the $\boldsymbol{\Sigma}$ denote the covariance matrices and we define $\mu_1 = \mathbb{E}[\mathbf{Y}_1] = \left[\mathbb{E}[X_i]\right]$, i.e., the matrix of the expected values of a univariate normal random variable $X_i$ for all $i$.
@Clarinetist: I don't have an example for you, but that seems very unlikely to me. Uncorrelated is such a weak statement I don't see why you should ever be able to promote it to independence unless one of the RVs is something like constant.
@MikeMiller Too lazy to type this out, but go here, plug in $\rho = 0$, and you can factor out the PDF $f_{X,Y}(x,y) = f_{X}(x) f_{Y}(y)$, hence independence
> However, it is possible for two random variables X and Y to be so distributed jointly that each one alone is marginally normally distributed, and they are uncorrelated, but they are not independent; examples are given below.
I gotta get Casella out
Ugh
@MikeMiller Here we go
Let $X$, $Y$ be independent $\mathcal{N}(0, 1)$ and define $$Z = \begin{cases} X, & XY > 0 \\ -X, & XY < 0\end{cases}\text{.}$$
It can be shown that $Z$ is normally distributed but $(Z,Y)$ is not
Well, that's a bummer. I wonder if there's a sufficient condition for joint normality...
I always like the idea of special geometries, things that only exist on very special types of manifolds; dimension constraints (like 7-dim) is particularly nice
this is also part of why it would be cool if there turned out to be an interesting Engel geometry!
@M.S.E I haven't explicitly tried that integral out, but note that the denominator is $a\cos^2 x + b \sin^2 x$. Now sub in $u = \pi/2 - x$ or something. You have to search for the symmetry in there, as @Ted indicated.
@Chris'ssistheartist TedShifrin said that alternatively without the conventional way of integrating $$\int_0^{\pi/2}\frac{1}{a\cos^2x + b\sin^2x}dx$$ by factoring $\cos x$ out, to find the R.H.S of the equation $I+J=\int_0^{\pi/2}\frac{1}{a\cos^2x + b\sin^2x}dx$ . He said to use more symmetry....
@Chris'ssistheartist With Balarka Sen advice, $x=\pi/2-x \implies I(a,b)=I(b,a)$
@Chris'ssistheartist However I dont know how to proceed with that.....
@Chris'ssistheartist is it $\displaystyle\int_0^{\pi/2}\frac{\cos^2x}{a\cos^2x + b\sin^2x}dx=\int_0^{\pi/2}\frac{\cos^2x}{a\cos^2x + b - b\cos^2x}dx$=$\int_0^{\pi/2}\frac{\cos^2x}{(a-b)\cos^2x + b }dx=\frac{1}{a-b}\int_0^{\pi/2}\frac{(a-b)\cos^2x}{(a-b)\cos^2x + b }dx=\frac{1}{a-b}\int_0^{\pi/2}1-\frac{b}{(a-b)\cos^2x + b }dx$
@Agawa001 I proposed to use $$I= \int_0^{\pi/2}\frac{\cos^2x}{a\cos^2x + b\sin^2x}dx $$ $$J= \int_0^{\pi/2}\frac{\sin^2x}{a\cos^2x + b\sin^2x}dx$$ and then we have the following system of equations: $1)$ $a I+b J =\pi/2$ and $2)$ $\displaystyle I+J=\int_0^{\pi/2}\frac{1}{a\cos^2x + b\sin^2x}dx$.
This way all is finalized immediately and elegantly.
After calculating the integral you have that $1)$ $a I+b J =\pi/2$ and $2)$ $\displaystyle I+J=\frac{\pi}{2\sqrt{a b}}$ from which you get the desired integral. Multiply the second equation by $-b$ and then add it to the first one.
(from here you get $I$)
More symmetry is effective when you have a nicer denominator, like $$\int_0^{\pi/2}\frac{\cos^2x}{\cos^2x + \sin^2x}dx$$.
off topic algebra. How can one see that the Galois group $G$ of the splitting field of $x^{49}-1$ over $\mathbb{Q}$ is cyclic? I have that this extension is Galois since $x^{49}-1$ is separable, and the extension is of degree $42$, but I am not sure why $G$ is cyclic.
@Chris'ssistheartist I dont have any difficulties, I succeeded with your advice :)
@Chris'ssistheartist But you asked if there was a shorter way, so some other user (Ted) said that you can use more symmetry and make it easier....so I was trying that and sharing with you.
@Chris'ssistheartist Just trying alternative methods you know.
@Chris'ssistheartist cause ted said you dont have to explicitly evaluate the integral to get the equation $I+J$ .
I might also think of Weierstrass substitution, but I didn't check this way.
@RandomVariable what do you think about the evaluation of the integral above? I used a system of equations (you can see it above), but not sure if there is a faster, simpler way.
@M.S.E Not sure you can skip that step with some trick.
@Chris'ssistheartist :) Because when I showed him my Integral evaluation :) He said there are other ways, and was encouraging by hinting to think more and use more of symmetrical properties. And on top of that Balarka Sen told me substitute $x=\pi/2-x $ which I(a,b)=I(b,a) so make use of that, but I didnt know what it was leading me to :)
@M.S.E It's a nice problem. Now I hope to note something interesting. You calculate the integrals of this type $$\int_0^{\pi/2}\frac{\cos x}{a \cos x + b \sin x}dx$$ in a similar fashion.
@Chris'ssistheartist really? This is really going to be a good question for me to really grasp this trick of yours. Because doing one time with help and one time alone is the best ;) Thanks.
@RandomVariable I was referring to $$\int_0^{\pi/2}\frac{\cos^2x}{a\cos^2x + b\sin^2x}dx$$, not to $\displaystyle \int_0^{\pi/2}\frac{1}{a\cos^2x + b\sin^2x}dx$.
@RandomVariable I used a system of equations to get that.
@Chris'ssistheartist Perhaps not as efficient, but you could multiply the top and bottom by $\sec^{2} x$, make the substitution $u=\tan x$, and then use partial fractions.
Yesterday I started in the same fashion, I did somewhere a tiny mistake and then I didn't want to continue since I considered the way is not that nice, but it was due to the mistake.
@RandomVariable yep, was about to do that. But @Chris'ssistheartist is right :) her's is easier, and for the first time Im using this way. Normally i always express after changing dx to du. $$=\int_0^{\infty}\frac{1}{(1+u^2)(a + bu^2)}du$$
:23854645 I don't think badges are reversed, but the next badge may be delayed until the qualification for two badges is met. That is how it was described to me.
@DanielFischer Let's say $f(z,a)$ has a finite number of simple poles on the positive real axis. If $|f(x,a)|$ and $|\frac{\partial}{\partial a} f(x,a)|$ are bounded in the intervals between the poles by integrable functions that are indepdent of $a$, is that enough to conclude $$\frac{d}{da} \, \text{PV} \int_{0}^{\infty} f(x,a) \, dx = \text{PV} \int_{0}^{\infty} \frac{\partial }{\partial a} f(x,a) \ dx? $$
@RandomVariable I have a question, might be a little bit personal. How do you and @Chris'ssistheartist @robjohn who are just really good at integrating come up there. Like see, overtime I see a new problem there is always some different substituting, etc (and I cannot guess what it is) . But you guys, my gosh like for example @Chris'ssistheartist gives what to do immediately. Like how do you guys do that? Has it been that way always?
@Chris'ssistheartist @RandomVariable @robjohn Like were you guys like me at the start? Not knowing what to do next and stuck alot and had to ask people? or am I not normal ?
@M.S.E Nobody is normal. And everybody starts not knowing what to do. It's practice, practice, and not to forget practice, that makes one see quickly what probably leads to success. Did I mention that it takes practice?
@M.S.E Absolutely. First of all, I only talk about me (robjohn did calculus already when he was 12 years old), I've never ever been gifted in mathematics or in anything else, but all I managed to do during the time was due to the extremely hard work, I worked very much and for long periods of times. With enough practice you'll do the same.
The key is simple: practice, practice, practice (it would be great if possible to work every day).
@RandomVariable If you stay away from the poles, you have no control over what happens when you shrink the holes. Then you can't deduce that you can interchange integration and differentiation.
@DanielFischer no I don't mean beginning. I know the theories, its not like I am talking about someone who does not know what calculus is, obviously one has to learn. But what I am referring to is substitutions and whats correct to do at the right and correct time
@M.S.E ;) Trust yourself, work hard, steadily if possible, accept that miracles won't happen over night, and you'll see that one day you'll do the same or better ;).
@M.S.E you don't need to have in mind comparisons with others, it might not make you feel good, but with yourself, to beat yourself every day in terms of performance.
(this is what I like to do, and if I see someone from which I can learn, or it seems more skillful than me, I'm glad because I have a opportunity to learn more)
@robjohn really you did calculus from the age of 12? Who or what made you introduced and interested? It can't be surely reading about it on the internet that introduced you, because interned is something recent in time.
Hard integrals will always be out there. In a way it's nice that we cannot do all immediately, you have the pleasure to think of stuff more, to find strategies to approach them, develop new tools, it's much pleasure in doing that. :-)
@Chris'ssistheartist last night I fell asleep with my laptop on :P and with the problems, wouldn't it have been super cool if I actually dreamt the solutions? :D
According to the Wikipedia article on unbiased estimation of standard deviation 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 indepe...
