if X is a continous variable and Y=X-mean(u), show that variance of Y = variance of X(sigma squared of X).?
345069
This can be shown for X as either continuous or discrete.  Let mu = E(X).\n\nVar(Y) = E(Y^2) - E(Y)^2\n= E[(X-mu)^2] - E(X-mu)^2\n= E(X^2 - 2Xmu + mu^2) - (E(X)-mu)^2\n= E(X^2) - 2muE(X) + mu^2 - (mu-mu)^2\n= E(X^2) - 2mu^2 + mu^2 - 0\n= E(X^2) - mu^2\n= E(X^2) - E(X)^2\n= Var(X)\n\nIf you need to be more specific than this, the replace E(X) = int(-inf to inf) xf(x)dx and E(X^2) = int(-inf to inf) x^2f(x)dx.  Everything should come out the same way.\n\nRemember that using substitution in the integral, let x = y + mu, the dx = dy, and the limits of integration would remain -inf to inf.
1043256
As you can tell from the various answers, your question is ambiguous.  If y is a function of x, then d(y^3)/dx = 3y²(dy/dx), using the chain rule, not + dy/dx as metaraison would have it.  If y is not a function of x, then dy/dx = 0 as if y were a constant.
717338
That depends on mu which would be a variable in only this situation.\nIt's the empty dairy which dictates if mu is constant or variable by the way.\n\nI'd like to go on and solve the problem for you but I do think that now,with a bit of effort you could do it yourself. As you will probably find you will experience a lot more pleasure if you do this.\nBest of luck \nIf you get stuck, dont be too shy in future to ask.>>>!~!!
740329
You will have to do an epsilon-delta proof.\nNotice that\n|f(x)-f(y)|=\n|x^3-y^3 -x+y|=\n|x-y|*|x^2+xy+y^2 -1|<=\n|x-y|*[|x|^2 +|x|*|y|+|y|^2 +1]\n<=148*|x-y|
900946
S.D. = ((Sigma xi^2 - (Sigma xi / n)) / (n-1))^1/2
391744
This is a simple two tailed test of the equality of means.\n\nLook this up in a stats book, but the Hypothesis that you are testing is Mean1=Mean2.
542781
The variables invovled can be correlated.  Below is a link for the multiviate normal.  Notice that the variance-covariance matrix Σ is not restricted to be diagonal; it can be a general matrix, it just has to be positive semi-definite.  Note that if |Σ| = 0, then the pdf does not exist.  Anyway, the link includes the pdf (in the positive definite case of course), bivariate case, and the conditional distribution, which is also normal.  I have been using this distribution a lot in my research.
887169
You're getting into logarithm.  If X/D = R^Y, then Y = log(X/D), where log is the logarithm in base R.
149140
Don't know Levey Jennings, but in general the variance is the sum of the squared deviations from the mean divided by the number of samples (for large data sets) or the number of samples minus 1 (for smaller data sets).  The standard deviation (SD) is then the square root of the variance.\n\nSo, if "m" is the mean and each sample is x1..n, the variance is:\n\n[(m-x1)^2 + (m-x2)^2 + (m-x3)^2 ... (m-xn)^2]/n\n\nThen, take the square root of the variance to get the standard deviation.
458036
If I understand your question correctly, this should help:\n\nTotal Variance = (Actual Quantity x Actual Rate) - (Budgeted Quantity x Budgeted Rate)\n\nIf you get a positive number, the variance is unfavorable and if you get a negative number, the variance is favorable.  \n\nHope this helps!
508493
I've found that the Wikipedia articles on probability distributions are top notch and contain enough information for you to solve this problem on your own.\n\nHint: solve the algebraic expressions that relate mean and variance so you can express p or n as algebraic relationship of one another and then do some substitions.\n\nAt first I thought you needed to know p or n up-front however I realize that you have all the info you need to solve this problem algebraically.\n\nGood luck with your homework.
229858
if you had an inverse variation, then x*y would always be the same number.\n\nif you had a direct variation, then x/y (or y/x) would always be the same number.\n\nonce you know which variation you have, then use the equations y=kx (for direct) and y=k/x (for inverse) to find the relation
295934
Covariance (or so I assume from the context) refers to the correlation between the two random variables. Because it's zero the total variance is just the sum of the two variances. \n\n(1-a)A and aB are the actual variables giving the returns for the B and A portions of the portfolio so the variances are (1-a)^2*(b^2) and a^2 * (2*b^2) (a^2 means a squared etc.). The total variance is just the sum:\n\n(1-a)^2*(b^2) + a^2 * (2*b^2) = b^2[1 - 2a + 3a^2]\n\nYou can see that if a=0 (all A) it's b^2 and if a is 1 (all B) it's 2b^2 as it must be. The minimum is found by differentiating w.r.t. a:\n\n6a - 2 = 0 ==> a = 1/3\n\nAt first it might seem that the portfolio should consist entirely of the stock with the lower volatility (A). However the admixture (2/3 A + 1/3 B) gives a lower volatility by virtue of the diversification provided by two investments instead of one.
798496
Let y = lnu where u = lnv where v = ln(6x)\n\ndy/dx = dy/du . du/dv . dv/dx\n= 1/u . 1/v . 6/6x\n= 1/ln(ln(6x)) . 1/ln(6x) . 1/x\n= 1/[xln(6x)ln(ln(6x))]
565556
This is counterexample is posted on Wikipedia; the link is below.  They don't explain all of the details, so I'll do that here.\n\nConsider the function f: R^2 -> R defined by\n\nf(x,y) = x y (x^2 - y^2) / (x^2 + y^2)\n\nI'm going to show that D_x (D_y f) is different from D_y (D_x f) at (x,y) = (0,0).\n\nThis function is continuous, because it is easy to check that lim_{(x,y) -> (0,0)} f(x,y) = 0.  Its derivatives are\n\nu(x,y) = D_x f(x,y) = y (x^4 + 4 x^2 y^2 - y^4) / (x^2 + y^2)^2\nv(x,y) = D_y f(x,y) = x (x^4 - 4 x^2 y^2 - y^4) / (x^2 + y^2)^2\n\nAgain, these functions are continuous, because lim_{(x,y) -> (0,0)} u(x,y) = lim_{(x,y) -> (0,0)} v(x,y) = 0.  Now to compute the mixed derivatives at (x,y) = (0,0), we need to compute the following limits.\n\nD_x (D_y f) @ (0,0) = lim_{h -> 0} ( v(h,0) - v(0,0) ) / h = lim_{h -> 0} h/h = 1\nD_y (D_x f) @ (0,0) = lim_{k -> 0} ( u(0,k) - u(0,0) ) / k = lim_{k -> 0} -k/k = -1\n\nHence these partial derivatives are *not* equal at (x,y) = (0,0) as claimed.\n\n-------------\n\nBy the way, I should point out that this gives a counter example to the famous statement\n\ndiv . curl (F) = 0\n\nIndeed, take F(x,y,z) to be that vector valued-function which is f(x,y) in the z-direction.  Then\n\ndiv . curl(F) = D_x (D_y f) - D_y (D_x f)\n\nIf you choose f(x,y) as above, then this expression is not zero at (x,y) = (0,0).
817419
Proof.\n\nsin : [a, b] → ℝ is a continuous differentiable function, so by the mean value theorem, ∃ c ∈ (a, b) such that\n\nsin'(c) = (sin(b) - sin(a)) / (b - a), i.e.\ncos(c) * (b - a) = sin(a) - sin(b).\n\nNow since ∀ x ∈ ℝ, -1 ≤ cos(x) ≤ 1, it follows that\n\n| sin(a) - sin(b) |\n= | sin(b) - sin(a) |\n= | cos(c) * (b - a) |\n= | cos(c) | * | b - a |\n≤ 1 * | b - a |\n= | a - b |. ∎
863028
Is that really the question?  Prob of a normal variable being greater than the mean?  It's just 0.5.  What's the real question, and what has Z to do with it?