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March 15

Normal distributed random variable mathematics

Suppose ((x, s)) and ((y, z)) are random variables with means x and y respectively, and standard deviations s and z respectively. Also consider a correlation coefficient p. Is there a page which lists the solutions to various functions f(((x, s)), ((y, z))) for example:

((x, s)) + ((y, z))
((x, s)) - ((y, z))
((x, s)) * ((y, z))
((x, s)) / ((y, z))
((x, s)) ^ ((y, z))
ln((x, s))
sin((x, s)) / cos((x, s)) / tan((x, s))

etc. and do the rules change with a probability distribution other than the normal distribution? —Preceding unsigned comment added by 203.22.23.9 (talk) 02:32, 15 March 2010 (UTC)[reply]

well, assuming that sample size is always equal for each RV (you're not, for instance, choosing 300 from X but only 30 from y) the mean of the new distribution should just be f(x,y). the standard deviation for the combined distribution will be a bit more of a problem to calculate. in the simplest cases (addition and subtraction) you can probably use the same combined SD formulas they use for two-sample z-tests and t-tests (square root of the sum of the variances, adjusted for differing sample sizes), but I am suspicious of using that approach even for multiplication and division, much less logarithms, power functions, and trig functions. I don't know if there's a general approach to this (and I'll confess some curiosity - anyone?) --Ludwigs2 03:24, 15 March 2010 (UTC)[reply]

I find the answer above confused. What is "f(x,y)"??? What sample size??? There's no sample size involved.

I'm going to call the random variables X and Y rather than x and y, since one shouldn't use the letter x to refer both to the random variable and to the expected value. Some information is missing above. It says X and Y or normally distributed and correlated, but it does not say whether they're jointly normally distributed (see bivariate normal distribution). For now, I'll assume that was intended. In that case we have a random vector $\left[{\begin{array}{c}X\\Y\end{array}}\right]$ with expected value $\left[{\begin{array}{c}x\\y\end{array}}\right]$ and covariance matrix $\left[{\begin{array}{cc}s^{2}&psz\\psz&z^{2}\end{array}}\right].$ The sum X + Y is equal to a product of matrices:

X+Y=[1,1]\left[{\begin{matrix}X\\Y\end{matrix}}\right]

and so we have a normally dstibruted random variable with expected value

E(X+Y)=E\left([1,1]\left[{\begin{matrix}X\\Y\end{matrix}}\right]\right)=[1,1]\left[{\begin{matrix}x\\y\end{matrix}}\right]=x+y

and variance

{\text{var}}(X+Y)=[1,1]\left[{\begin{array}{cc}s^{2}&psz\\psz&z^{2}\end{array}}\right]\left[{\begin{array}{c}1\\1\end{array}}\right]=s^{2}+2psz+z^{2}.

A similar method works for the difference. The quotient has a Cauchy distribution; that one's a bit more work. The logarithm is problematic because the ranges of these random variables include negative numbers and those don't have logarithms. Michael Hardy (talk) 03:50, 15 March 2010 (UTC)[reply]

The OP said pretty clearly what f is, what x is and what y is. I'm not sure what the source of confusion is. Rckrone (talk) 05:26, 15 March 2010 (UTC)[reply]

OK, I see that the OP did say what f(x,y) is. But that makes the proposed answer wrong. The mean is not f(x,y). What makes you think it is? Michael Hardy (talk) 16:12, 15 March 2010 (UTC)[reply]

The "sample size" bit is pretty confusing. -- Meni Rosenfeld (talk) 09:19, 15 March 2010 (UTC)[reply]

Actually negative numbers do have logarithms, but they're complex.--220.253.247.165 (talk) 05:38, 15 March 2010 (UTC)[reply]

The mean is f(x, y), that's easy, I wouldn't know if there's a general form like that for finding the standard deviation. Does anyone know? --220.253.247.165 (talk) 05:53, 15 March 2010 (UTC)[reply]

The mean is definitely not

f(x,y)

. For

f(X,Y)=XY

this requires the variables to be uncorrelated; for more general functions even that's not enough. It's not even true in the one-dimensional case:

\mathbb {E} [f(X)]

is not

f(\mathbb {E} [X])

. For example,

\mathbb {E} [X^{2}]\neq \mathbb {E} [X]^{2}

for any variable with nonzero variance. -- Meni Rosenfeld (talk) 09:19, 15 March 2010 (UTC)[reply]

See Multiset#Cumulant_generating_function.

Using the plus-minus sign notation ((x, s))=x±s for a random variable having mean value x and standard deviation s, the simple formulas are

(x±s)+y = (x+y)±s

(x±s)−y = (x−y)±s

(x±s)*y = xy±sy

(x±s)/y = x/y±s/y

These formulas are easy to remember because they look like the associative and distributive rules of algebra. Also assuming independence you have

(x±s)+(y±z) = (x+y)±√(s²+z²)

(x±s)−(y±z) = (x−y)±√(s²+z²)

The more complex formulas do not exist because the mean value and the standard deviation of a function generally depend not only on the mean values and the standard deviations of the arguments, but on further details of the distributions. Bo Jacoby (talk) 08:47, 16 March 2010 (UTC).[reply]

Representation of Haar scaling function in terms of Haar wavelet

The set of Haar wavelets $\{t\mapsto \psi _{n,k}(t)=\psi (2^{n}t-k);n\in \mathbb {N} ,0\leq k<2^{n}\}.$ is a complete orthonormal system of the unit interval ... what is a representation of the Haar scaling function (the function that is 1 on the unit interval and 0 everywhere else) in this basis? I have been thinking about this for a while but for some reason am at a total loss. Thanks, 98.202.53.240 (talk) —Preceding undated comment added 04:56, 15 March 2010 (UTC).[reply]

I think the theorem is actually that these mother wavelets, with the scaling function, are an orthonormal basis. This could be a mistake in our article.

Just the mother wavelets obviously can't represent the scaling function, since any

f(t)=\sum _{n=0}^{N}\sum _{k=0}^{2^{n}-1}a_{n,k}\psi _{n,k}(t)

satisfies

\int f(t)\,dt=0

, so

\int (f(t)-\phi (t))^{2}\,dt=\|f\|^{2}+1\geq 1

, so linear combinations are not arbitrarily close to

\phi (t)

. -- Meni Rosenfeld (talk) 09:37, 15 March 2010 (UTC)[reply]

Thanks, that makes sense. Not sure what to do now though :-/ --98.202.53.240 (talk) 01:44, 16 March 2010 (UTC)[reply]

Do about what? -- Meni Rosenfeld (talk) 08:15, 16 March 2010 (UTC)[reply]

Cases where Euclid's first common notion doesn't apply

What are some cases where you can't say "Things that equal the same thing also equal one another."?20.137.18.50 (talk) 14:38, 15 March 2010 (UTC)[reply]

I would say no, by definition. A relation has to be an equivalence relation (which requires it to be transitive, ie. A=B, B=C => A=C) to be considered a form of equality. --Tango (talk) 15:04, 15 March 2010 (UTC)[reply]

Sometimes, though, the notation and language for equality are used for what is not really a form of equality. For example, the Landau notation uses a very specific notion of equality which is not symmetrical. Thus

x=O(x)

and

2x=O(x)

but

x\neq 2x

. -- Meni Rosenfeld (talk) 15:13, 15 March 2010 (UTC)[reply]

True. I would say that was an abuse of notation (in fact, it is given as an example in that article), rather than a strange definition of equality. People use the equals sign (and the word "is") to refer to something that is not equality, rather than it being a strange type of equality. The distinction is rather pedantic, admittedly. --Tango (talk) 15:38, 15 March 2010 (UTC)[reply]

I think Landau notation uses "=" and "is" in a similar way to the predicative use of "is" in normal English. Thus Meni's example is of the same sort as "My house is blue. The sea is blue. Thus my house is the sea." Algebraist 16:06, 15 March 2010 (UTC)[reply]

Indeed, the sign "=" in Landau notation, like e.g. in "f=o(g)", "o(g)=O(h)" stands respectively for "∈" and "⊆"; "O(f)" "o(f)" should be thought as set of functions (actually modules). --pm a 20:48, 15 March 2010 (UTC)[reply]

How about where multiple values exist, and only one is listed ? Such as:

 $2={\sqrt {4}}$ 
 ${\sqrt {4}}=-2$ 
 $2=-2$

StuRat (talk) 16:05, 15 March 2010 (UTC)[reply]

Your 2nd line is simply wrong. The square root symbol is defined as the positive square root. If multiple values exist, you shouldn't really say it equals one of them, you should say something like "4^1/2=+/-2". --Tango (talk) 16:14, 15 March 2010 (UTC)[reply]

The use of the symbol aside, I'm sure you see what I meant. StuRat (talk) 19:36, 15 March 2010 (UTC)[reply]

Yes, but even if you had said 4^1/2 I still wouldn't accept that 4^1/2=-2 was a true statement. (Since 4^1/2=+/-2, as I already said.) --Tango (talk) 20:09, 15 March 2010 (UTC)[reply]

To generalize, StuRat is stating that a function that maps a space onto a smaller subspace causes a problem. If F(x) maps a space onto a smaller subspace, it is obvious that it maps more than one value to a common value. Let's assume F(1)=10 and F(2)=10 for this function. Then, if is common to say F(1)=10, F(2)=10, F(1)=F(2). But, when you have a function that is written as a symbol (like the square root symbol), you get people writing something like 1=*10, *10=2, 1=2... where 1=*10 is a symbolic way of writing F(1)=10. So, again, it is a bit of abuse of notation. If you are stuck with "S(x)=square root of x" being the only notation, it is harder to say 2=S(4), S(4)=-2, 2=-2. -- kainaw ™ 20:37, 15 March 2010 (UTC)[reply]

To say that more precisely, functions that aren't injective can't have inverses. Rckrone (talk) 21:16, 15 March 2010 (UTC)[reply]

Exactly. Kainaw's * symbol is denoting the inverse of F, which doesn't exist (or, if you prefer, isn't a function). You can't then apply the usual rules for manipulating functions and expect it to work. --Tango (talk) 21:25, 15 March 2010 (UTC)[reply]

crisis in geometry

what are the crisis in geometry? recent trends in crisis in geometry? —Preceding unsigned comment added by Sunil moktan (talk • contribs) 15:44, 15 March 2010 (UTC)[reply]

I don't think there are any crises in geometry... could you explain a little more about what you want? --Tango (talk) 16:15, 15 March 2010 (UTC)[reply]

Googling "crisis in geometry" throws up a lot of references to this essay, but this talks about a crisis in art, not in mathematics. Gandalf61 (talk) 16:20, 15 March 2010 (UTC)[reply]

The discovery of irrational lengths was a crisis to some, as was the development of Non-Euclidean geometry. <joke> Also, Crysis does some fairly heavy geometric calculations. </joke> -- Meni Rosenfeld (talk) 18:41, 15 March 2010 (UTC)[reply]

Maybe the OP wants to know what some of the open problems and current areas of research are in geometry. Rckrone (talk) 21:32, 15 March 2010 (UTC)[reply]

There was a so-called foundational crisis of mathematics. Do you mean that? 66.127.52.47 (talk) 01:24, 16 March 2010 (UTC)[reply]

Here's an improved link: Foundations_of_mathematics#Foundational_crisis.StuRat (talk) 01:47, 16 March 2010 (UTC)[reply]