If the moment generating function of a random variable X is given by?
964589
to find the mean and variance of a MGF you need to find the moments.  To find the first moment, take the first derivative and then set t=0.  This would be (3 + 16t)*e^(3t + 8t^2).  With t=0 this equals 3.  The second moment is equated from the second derivative.  This is 16*e^(3t + 8t^2) + (e^(3t + 8t^2))*(3+16t)^2.  With t=0 the second moment equals 16 + 9 = 25.  \n\nThe mean is the first moment when t=0.  This equals 3.  \n\nThe variance is the second moment minus the first moment squared.  This equals 25 - 9 = 16.
206380
basically, you generate a list of random numbers (more about random number generation below), then use those numbers to calculate some output function.\nThen, you can have an "empirical distribution" about your output function.\nExample:\nyou can compute the present value of a financial project, GIVEN the values of, say, interest rates , sales level and inflation.\nYou also know that these three variables distribute normally and are correlated.\nThen you can generate a random scenario (one set os observations for each variable) and compute present value for it. You repeat the process a lot of times.\nSince you have a PV for each scenario you can observe its probability distribution.\nThat's it about Montecarlo Simulation.\nAbout Random number generation:\nif you know its cummulative probability distribution function (F(X)), then you can generate uniformly distributed random number in then (0,1) interval; let that be P.\nThe F-1(P) distributes with PDF F,where F-1 is teh inverse function of the cummulative function.\nThis method can be easily expanded to multivariate simulation using independt uniform variables, and the cholesky decomposition of the covariance matrix.\nNow, a note of care:\n\nMontecarlo methods can give the sensation that you can observe the distribution of a variable even if you know nothing about it. But, at the end, it's nothing more than a METHOD for computing PDF's  that are not easily expressed as functions.\n\nFinally, a note about computer power.\nMontecarlo methods are power hungry.Thus, using them to simulate PDF's of variables can indeed be expressed analytically is a complete waste of computin power and human time.
217977
f(E) = 1/(exp{(E-Ef)/kT}+1)
821550
if f(x) = ∫[t = x to 0] 1/ (1 + t + t²) dt\n\nf'(x) = -1/ (1 + x + x²)\n\nf''(x) = (1 + 2x)/(1 + x + x²)²\n\nFor concave UP f''(x) > 0\n\nSo 1 + 2x > 0 (as (1 + x + x²)² > 0 for all x)\n\nie x < -1/2
794682
Suppose X is a continuous random variable whose density function f is defined by f(x) = x/2 whenever 0<x<2 and f(x) = 0 otherwise. Then,\nE[X] = integral (x * f(x), x=0..2 ) dx\n     = integral (x * x/2, x=0..2 ) dx\n     = (1/2) integral (x^2, x=0..2 ) dx\n     = [(1/6)*x^3, x=0..2] \n     = (1/6)*2^3 - (1/6)*0^3 \n     = (8/6)\n     = (4/3).\n\nNext, \nE[ | X - E[x] | ] = E[ | X - 4/3 | ] \n     = integral ( | x - 4/3 | * f(x), x=0..4/3 ) dx + integral ( | x - 4/3 | * f(x), x=4/3..2 ) dx\n     = integral ( - (x - 4/3) * f(x), x=0..4/3 ) dx + integral ( (x - 4/3) * f(x), x=4/3..2 ) dx\n     = integral ( (4/3 - x) * x/2, x=0..4/3 ) dx + integral ( (x - 4/3) * x/2, x=4/3..2 ) dx\n     = integral ( (2/3)x - (1/2)x^2), x=0..4/3 ) dx + integral ( (1/2)x^2 - (2/3)x, x=4/3..2 ) dx\n     = [(1/3)x^2 - (1/6)x^3, x=0..4/3 ] dx + [(1/6)x^3 - (1/3)x^2, x=4/3..2]\n     = ( (1/3)(16/9) - (1/6)(64/27) - 0 + 0) + ( 8/6 - 4/3 - (1/6)(64/27) + (1/3)(16/9))\n     = ( 16/27 - 32/81 ) + ( 16/27 - 32/81 )\n     = 2(16/27 – 32/81)\n     = 32/81.\n\n--- \n\nSome notation used:\nintegral (f(x), x = a..b) dx means we evaluate the definite integral of f(x) bounded by x = b and x = a.\n\n[ F(x), x =  a.. b ] means we evaluate the difference, F(b) - F(a).\n.
733701
let us denote the marginal density of X_i by f_i, that is the density of X_1 is f_1.\nlet us denote the density of the triplet (X_1, X_2 X_3) by f.\n\nas X_1, X_2, X_3 are stochasticaly independent,\n\nf(x_1, x_2, x_3) = f_1(x_1)  * f_2(x_2) *  f3(x_3)\n\nlet g be a bounded borelian function on R ^3\n\nset:\nphi(x_1, x_2, x_3) = (x_1 + x_2 + x_3) ^ 2 / 6\n\nE(g o phi (X_1, X_2, X_3)) = &#92;int g o phi(x_1, x_2, x_3) f(x_1, x_2, x_3) dx_1 dx_2 dx_3\n\n = &#92;int g( ( x_1 + x_2 + x_3) ^ 2 / 6) f(x_1, x_2, x_3) dx_1dx_2dx_3\n\nthen you may conclude using Fubini theorem
363348
Are we solving for x?
364344
emmmmmm\n\nd. its not a fuction of x in terms of y explicitly. It needs to be rearranged
669001
e^f(x) = x is the best answer i can give you.
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.
790936
x≠０\ncuz if x=0, (x-7)/x^3 is meaningless
228750
x xxxxx x\n    x----->\n\n<--x    x x--->\n    x
1048409
It can be anything.  If you follow the sequence to (x-x) you will get an answer of 0 for the entire equation.
920901
You can get it the following way:\n\nThe cdf of Xbar is \nF(y) = P((X1+X2+...+Xn)/n < y) = P(X1+X2+...+Xn < n*y) = \nP(sqrt(n)*(X1+X2+...+Xn-mu)/ sigma < sqrt(n)*(n*y-mu)/sigma) = Phi(sqrt(n)*(n*y-mu)/sigma), where Phi(.) is the cdf of the standard normal distribution.\n\nDifferentiating this with respect to y yields the pdf of Xbar:\nf(y) = phi(sqrt(n)*(n*y-mu)/sigma) *n*sqrt(n)/sigma, where phi(.) is the pdf of the standard normal distribution.\n\nAs a consequence of the CLT, for large values of n, this does not depend on the distribution of the original X1,X2,...,Xn, only on its mean and standard deviation. However, the CLT is only an approximation.\n\nIf you'd like to get the accurate pdf of Xbar, you need to use the convolution formula for the pdf of the sum of IID random variables. This is pretty nasty, so I'd rather suggest that you use moment generating functions or characteristic functions instead. (It is very simple to compute the mgf and the chf of the sum of IID random variables; in fact, the most well-known proof of the CLT is also based on this technique.)
726082
Yes since ∫f(x) dx = ∫x dx\n= ½x² + c where c is ANY constant.
952174
That is the central limit theorem.  Each distribution can be uniform, but the distributions themselves are normally distributed in their range, their value, their frequency.\nRead the attached link for more details.
514301
Did you mean stochastic? It's a random variable quantity.