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March 24

Expected value of variance

The article Bias of an estimator states:

\operatorname {E} (S^{2})={\frac {n-1}{n}}\sigma ^{2}.

Why is that so? Is that really the case for all distributions? — Sebastian 05:02, 24 March 2007 (UTC)[reply]

Is that just because the variance is calculated by summing the square deviations and then dividing by (n-1), so you have to multiply by (n-1)/(n) to correct for that, and to make it as if you'd divided by n in the first place? Perhaps I'm not understanding the question - I'm no stats expert, and I've never understood why the formula for variance has that (n-1) in the denominator.

Actually, the paragraphs at the end of the section Variance#Introduction might answer your question and mine. I'll have to stare at them for a bit longer to be sure... -GTBacchus^(talk) 05:35, 24 March 2007 (UTC)[reply]

Thanks for that pointer. Yes, that's a nice explanation in Variance#Introduction, but it's a bit handwaving. What I meant, though, was: The deduction in Bias of an estimator looks like it is meant to be a proof why we divide by n-1. Therefore, we can't assume that we know that already. Either that part of the article is not meant as a deduction or we're missing a basic step. — Sebastian 06:04, 24 March 2007 (UTC)[reply]

The statement in Bias of an estimator quoted above is not a deduction, and no proof is given, but if you accept that it is true, it shows that S² is a biased estimator of σ², and further that it can be made unbiased by multiplying it by n/(n−1) if n > 1. The statement holds for all distributions with finite mean and variance. It is a tedious but otherwise elementary exercise to show that it holds. The essential step is to show that

\operatorname {E} (({\overline {X}}-\mu )^{2})={\frac {1}{n}}\sigma ^{2}.

This can be obtained by expanding

{\overline {X}}

into an averaged sum, bringing μ into the summation, expanding the product of two sums into a sum of n² products, distributing the E(•) operator as much as possible, using the independence of X_i and X_j for i ≠ j, and simplifying. --Lambiam ^Talk 12:29, 24 March 2007 (UTC)[reply]

Thanks for the great answer! That answers my question. The reason why I expected a proof was that the Standard deviation article says "The reason for this definition is that s² is an unbiased estimator ...". The word "reason" raises expectations; not sure if we should reword that. — Sebastian 22:51, 24 March 2007 (UTC)[reply]

Angular size

Hi. I know I have made many formulas, but this one always puzzles me. Is there an easy way to determine, or at least estimate, how far away something is by observing its apparent size and knowing its size, or how large something is by its apparent size and distance? I'm no expert in math, but I have created formulas using logic. So, please tell me an easy way to calculate, and any errors in my hypopethized formula. $S*(60/A)/I?O=D$ and $D*(A/60)*I?O=S$ , where:

S is the size of the object
D is the distance of the object
A is the apparent size of the object, eg. a fist held at arm's length ≈ 10°
I is the length of one side of an equilaterial triangle divided by "the length of a line that would go from the center of the side of the triangle to the corner of the triangle perpendicularly so that it could only end up on one of the points so that the line formed does not overlap the line the first point was on"
O is the number of times, for example, an isoceles or equilateral triangle is split down the middle, it's the number of times the resulting angles is more than the original angle, it's hard to explain, and I'm not sure if you would multiply or divide by these for each A stroHur ricane 00 1^{(Talk+Contribs+Ubx)} 19:33, 24 March 2007 (UTC)[reply]

If an object is sufficiently far away, its diameter (I am guessing this is what you mean by size) is approximately equal to its distance times the angle (in radians) that its diameter takes up in your field of view, so I guess you can say something like D = S / A, where A is in radians. --Spoon! 21:27, 24 March 2007 (UTC)[reply]

Aslo, if you could try to see if my parallax formula is correct. $D=Q/W*F$ and $S=W/Q*G$ , where:

S and D are the same as above
Q is the angle of the two points of view with the central point between the observer and the object
W is the angle with the two sides of the object with the central point
F is the distance between the middle of the two observing points and the central point
G is the distance between the two observing points

I hope you understood me, I had to ponder very intensively. If my logic and reasoning are off, please say so. Thanks. A stroHur ricane 00 1^{(Talk+Contribs+Ubx)} 19:33, 24 March 2007 (UTC)[reply]

AstroHurricane - my first reaction is that you need to be more precise in your definition of terms. Let's take some examples:

When you say that A is "the apparent size of the object", how is this "apparent size" measured ? Is A an angle ? If so, is it measured in radians or degrees ?
I think you are trying to define I as the ratio of one side of an equilateral triangle to its perpendicular height. If you are, then a much more precise and economical way of expressing this is to say that I is the cosecant of 60 degrees.
Your definition of O is so obscure that I could not make any sense of it at all.
In your second formula, D is the distance of the object - but distance from where ? Distance from the central point ? Or distance from one of the observing points ?

I tried to get past your unusual definitions to give you some constructive feedback on the formulas themselves, but I am afraid I was so uncertain about the meaning of many of your terms that I eventually gave up. Understanding your definitions would be much easier of you provided some diagrams (like the one in triangulation, for example). It would also help if you learned and used some of the standard terminology of geometry - you may need to learn some trigonometry as well. Bottom line is that it is very difficult for anyone to evaluate your work if they have to hack through a thicket of oddly worded, imprecise and non-standard definitions first. Gandalf61 09:32, 25 March 2007 (UTC)[reply]

Apparent size (also known as "Angular size" or "Angular diameter" – which for unclear reasons is a separate article) is a well-known and well-defined concept, and the example with 10° makes it clear the questioner is measuring the angle in degrees (which is quite usual but gives more complicated formulas for the relation between true size, distance, and angular size). I also don't understand the definition of O, nor those of Q, W, F and G in the next part about parallax. Also, I see a question mark between I and O in the formulas. To work for angular sizes measured in degrees, it should be the case that I?O = π/3. --Lambiam ^Talk 12:43, 25 March 2007 (UTC)[reply]

Hi. Sorry if you misunderstood me, I will clarify. D, in the parallax formula, is the mean distance to the object from the two points, and also the distance to the object from the point in between the two observing points; D is the average mean of those two. A is supposed to be in degrees, for example a fist held at arm's length ≈ 10°. I put a question mark because I wasn't sure whether it was to multiply or to divide, I had trouble figuring it out, and there are no parentheses near the I or the O. Generally, I≈1/0.866, and O≈~vaugely, with or without division with 1, about 1.1, this is hypopethical. It is what discrepancy needed to conterbalance any errors. Now, to clarify the parametres, these are vauge or approximate. You know how there's two observing points, a central point, and two sides of the object in the parallax formula? This forms an approximate x shape. Now let's say the side toward the observer is the bottom of the x, and the side toward the top is the side near the object is the top. Q is the angle, in degrees or otherwise, at the bottom of the X. W is the angle at the top of the X. F would be if there was a horizontal line at the bottom of the X; F is the distance between the centre of that line and the centre of the X. G is the length of the new line mentioned at f, the distance between the two points at the bottom of the X, the horizontal line at the bottom of the X. I'm sorry if this made it more confusing, please understand that I'm not an expert mathematician. Is it now possible to see if my formula has any errors? If you understand, please answer. Thanks. A stroHur ricane 00 1^{(Talk+Contribs+Ubx)} 15:23, 25 March 2007 (UTC)[reply]

No, that's not any clearer - sorry. Let's take W and Q, just for starters. From your description of them being at the "top" and "bottom" of the "X", it seems as if they might be vertical angles (or opposite angles as we would say in UK). But if this were the case then W and Q would be equal, W/Q would always be 1, and your formulae would be obviously incorrect. Like I said before, you really need to provide a diagram to show which angles you actually mean here. Gandalf61 16:37, 25 March 2007 (UTC)[reply]

Sorry about that, now I'm confused. They may or may not be vertical angles, but I'm thinking that if the central point was wherever the lines intersected, it may cause problems, and if it was in a definite place, the lines would be bent at the central point, it may also cause problems. Let's assume it either was or wasn't a vertical angle. If it was, then you said it's incorrect. If it was, it would be a bent crooked X, would that work? I never meant it was a perfect x, it could probably be bent, although that may falsify the formula and confuse its relation upon itself. Anyway, you asked for a diagram, this would only work if your browser has the correct font, I don't really want to upload just yet. Here it is. Remember it's approximate and may not be accurate or vertical angles. This is only for the parallax.

                object
                 |
                 |
                 V
              \---------/ 
               \   |   /
                \  |D /
                 \ | /
                  \|/<--W
               Q->/|\ <-- central point
                 /F|D\
                /-----\
                   G
                ^    ^
                |    |
            observing points

I hope this helps to clarify. If you could answer, please do so for both. The article on triangulation I've read, but I asked if my formulae were triangulation or trigonometry, and nobody seems to have answered. Please answerif possible. Thanks. A stroHur ricane 00 1^{(Talk+Contribs+Ubx)} 22:04, 25 March 2007 (UTC)[reply]

Triangulation is an application of trigonometry, which incidentally means "measurement of triangles". Yes, you're trying to use trigonometry to triangulate the distance to something, but I'm not sure you've succeeded. A question on the parallax thing - what do you mean by "crooked"? If the lines are bent, it seems to defeat the purpose of the thing. Black Carrot 23:08, 25 March 2007 (UTC)[reply]

Hi. Sorry for the confusion. If the lines are straight, it also seems to defeat the purpose, see above. If you can figure it out, please try to see whether or not my formulae are correct, and implement any corrections if nessecary. If nobody can figure it out or it's not even close to being correct, please show me a place to find the correct formula, if there is one. I'm not really sure if I fully understand the process on the triangulation article, though, so if possible, please at least explain how triangulation works, the formulas for it, and if there is an easy way to calculate by triangulation using an average calculator. Ialso asked a few questions on the traingulation talkpage, but I haven't yet seen a response yet. Thanks. A stroHur ricane 00 1^{(Talk+Contribs+Ubx)} 00:20, 28 March 2007 (UTC)[reply]

stellar brightness

Hi. I wanted to ask if this formula is correct. this is an easier one. $B=D^{3}*0.5/E^{2}$ This formula is only used to compare brightness, and assumes that both stars are the same density, same brightness per unit of mass and volume, and same type. You must use the same units. This is where:

B=comparitive brighness
D=diametre of star
E=distance of star from observer

Could someone answer? Thanks. A stroHur ricane 00 1^{(Talk+Contribs+Ubx)} 19:49, 24 March 2007 (UTC)[reply]

This might be a question better asked at the science desk. --Ķĩřβȳ ♥♥♥Ťįɱé Ø 23:25, 24 March 2007 (UTC)[reply]

If you only want to compare, the factor 0.5 is superfluous. --Lambiam ^Talk 00:30, 25 March 2007 (UTC)[reply]

The equation looks like it would be correct if your assumptions were correct, but your assumptions look like they would not normally be met for most pairs of stars. For one thing, assuming the two stars in question have the same density is quite unlikely if the stars have a different mass. In general, the greater the mass of the star, the stronger the compressive force of gravity will be, so the greater the density is likely to be. If you look at the Main Sequence Star Properties table on this page[1], you'll see that stars along the middle part of the main sequence have a mass that's very roughly proportional to D, not to D³ like would be required if the density remained constant. More importantly for the purposes of this calculation, you're assuming that the star's luminosity is proportional to the star's mass or volume, neither of which is normally the case. The luminosity is actually proportional to D²T⁴, where T is the star's surface temperature. [2] (The brightness is proportional to D²T⁴/E².) Your assumption would only work as some kind of approximation if D were roughly proportional to T⁴. A glance at the Main Sequence Star Properties table will show that that's not even close to being accurate in general. MrRedact 06:12, 25 March 2007 (UTC)[reply]