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Normal Equations 1The result of this maximization step are called the normal equations b 0 and b 1 are called point estimators of 0 and 1 respectively X Y i = nb 0 b 1 X X i X X iY i = b 0 X X i b 1 X X2 2This is a system of two equations and two unknownsQED Definition An element y in B is called a complement of an element x in B if xy=1 and xy=0 Theorem 2 For every element x in B, the complement of x exists and is unique Proof Existence Let x be in B x' exists because ' is a unary operation X' is a complement of xE−λ = λ X∞ k=0 λk k!
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E^x-y can be written as
E^x-y can be written as-2 0 ≤ X ≤ 1 EY X = X − 1 2 1 < X ≤ 2 (b) Let g(x) be the estimate from part (a) Find Eg(X) and var(g(X)) g(X) is a derived random variable that is defined as g(X) = 1 2, 0 ≤Y 2 − e x y − e x = 0 This is a second degree polynomial in y;
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Var b k Var y k k Cov y y xx Cov y y y y s s s s The variance of b0 is 2 Var b Var() ()2 (,)011 y xVarb xCovy b First, we find that 111 1 01 1 (,) () 1 ()( ) 1 0000 0 ii i ii iiii i ii i i i Cov y b E y E y b E b Ecy Eccxc n n So 2 2 0 1 xx x Var b ns Covariance The covariance between b0 and b1 is 01 1 1 2 (,) (,) () xx Cov b b Cov y bIntegrate x^2 sin y dx dy, x=0 to 1, y=0 to pi;Y 1(x) = e3x and y 2(x) = e−2x These are linearly independent and therefore the general solution is y cf(x) = Ae3x Be−2x The equation k2 −k −6 = 0 for determining k is called the auxiliary equation Task By substituting y = ekx, find values of k so that y is a solution of d2y dx2 −3 dy dx 2y = 0
Answer (1 of 3) First, I request you to kindly read the chapter on Expectation from any basic book on Statistics There you will find the following (i) If X is a random variable taking the values x_i ( i=1,2,3,n) with corresponding probabilities as p_i(i=1,2,3,n) then Expectation of x isIt is also, of course, a function ofClick here👆to get an answer to your question ️ If x^y = e^x y , then dydx is equal to Join / Login >> Class 12 >> Maths >> Continuity and Differentiability >> Derivatives of Implicit Functions If x y y x = a b then find that d x d y
Proof lnexy = xy = lnex lney = ln(ex ·ey) Since lnx is onetoone, then exy = ex ·ey 1 = e0 = ex(−x) = ex ·e−x ⇒ e−x = 1 ex ex−y = ex(−y) = ex ·e−y = ex 1 ey ex ey • For r = m ∈ N, emx = e z }m { x···x = z }m { ex ···ex = (ex)m • For r = 1 n, n ∈ N and n 6= 0, ex = e n n x = e 1 nx n ⇒ e n x = (ex) 1 • For r rational, let r = m n, m, n ∈ NX and Y, ie corr(X,Y) = 1 ⇐⇒ Y = aX b for some constants a and b The correlation is 0 if X and Y are independent, but a correlation of 0 does not imply that X and Y are independent 33 Conditional Expectation and Conditional Variance Throughout this section, we will assume for simplicity that X and YLet X and Y be two discrete random variables Then E (aX bY) = aE (X)bE (Y) for any constants a,b ∈ R Note No independence is required Proof E (aX bY) = P x,y (axby)p(x,y) = a P x,y xp(x,y)b P x,y yp(x,y) P x p(x,y) = p(y) x = a P x xp(x)b P y yp(y) = aE (X)bE (Y) Example Binomial distribution Let X ∼ Bin(n,θ) Then X = X 1
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7b E(a ± X) * b = (a ± E(X)) * b 8 E(X Y) = E(X) E(Y) (The expectation of a sum = the sum of the expectations This rule extends as you would expect it to when there are more than 2 random variables, eg E(X Y Z) = E(X) E(Y) E(Z)) 9 If X and Y are independent, E(XY) = E(X)E(Y)The nature of the underlying relation is Y = a e b x, where alpha and beta are parameters of the relation To get this relation in linear model form, one transforms both sides of the equation to obtain ln(Y) = ln(a e b x) = ln(a) ln(e b x) = ln(a) b x = b 0 b 1 x In linearized form b 0 = ln(a) and b 1 = bE(aX bY c) = aE(X)bE(Y)c for any a,b,c ∈ R 1 Let's use these definitions and rules to calculate the expectations of the following random variables if they exist Example 1 1 Bernoulli random variable 2 Binomial random variable 3 Poisson random variable 4 Negative binomial random variable
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(b) P(Y 3) = F(3) = R 3 2 x 6 dx= x 2 12 i x=3 x=2 = 3 12 2 12 = 5 12 requires (i) summation (ii) integration and is a value of a (i) probability density function (ii) cumulative distribution function which is a (i) stepwise (ii) smooth increasing function (c) E(Y) = (i) P yf(y) (ii) R yf(y)dy (d) Var(Y) = (i) E(X2) 2 (ii) E(Y2) 2 (e) M(t(b 0;b 1) E (Y (b 0 b 1X))2 (5) This is a function of the complete distribution, so we can't get it from data, but we can approximate it with data The insample, empirical or training MSE is MSE\(b 0;b 1) 1 n i=1 (y i (b 0 b 1x i)) 2 (6) Notice that this is a function of b 0 and b 1;Expected Value and Standard Dev Expected Value of a random variable is the mean of its probability distribution If P(X=x1)=p1, P(X=x2)=p2, n P(X=xn)=pn E(X) =
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Department of Computer Science and Engineering University of Nevada, Reno Reno, NV 557 Email Qipingataolcom Website wwwcseunredu/~yanq I came to the USGraph y=e^ (x) y = e−x y = e x Exponential functions have a horizontal asymptote The equation of the horizontal asymptote is y = 0 y = 0 Horizontal Asymptote y = 0 y = 0E(x,t) = E max cos(kx ωt φ), B(x,t) = B max cos(kx ωt φ) E is the electric field vector, and B is the magnetic field vector of the EM wave For electromagnetic waves E and B are always perpendicular to each other and perpendicular to the direction of propagation The direction of propagation is the direction of E x B Let the fingers of your right hand point in the direction of E
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Integrate x/(x1) integrate x sin(x^2) integrate x sqrt(1sqrt(x)) integrate x/(x1)^3 from 0 to infinity;The fitted regression line/model is Yˆ = X For any new subject/individual withX, its prediction of E(Y)is Yˆ = b0 b1X For the above data, • If X = −3, then we predict Yˆ = − • If X = 3, then we predict Yˆ = • If X =05, then we predict Yˆ = 2 Properties of Least squares estimators It is usually best to see how we use these two facts to find a potential function in an example or two Example 2 Determine if the following vector fields are conservative and find a potential function for the vector field if it is conservative →F = (2x3y4 x)→i (2x4y3 y)→j F → = ( 2 x 3 y 4 x) i → ( 2 x 4 y 3 y) j →
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E−λ = λ The easiest way to get the variance is to first calculate EX(X −1), because this will let us useB r(x) = fy2X d(x;y) 0 and center x2Xas the set of points whose distance from xis less than or equal to r, B r(x) = fy2X d(x;y) rg The term \ball" is used to denote a \solid ball," rather than the \sphere" of points whose distance from the center xis equal to r Example 1311Let us check directly that E is a base for a topology Suppose that x,y∈X and ε,δ>0If z∈B(x,δ)∩B(y,ε),then B(z,α) ⊂B(x,δ)∩B(y,ε) (102) where α=min{δ−d(x,z),ε−d(y,z)},see Figure 102 This is a formal consequence of the triangle inequality For example let us show that B(z,α) ⊂
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In this case if b or x is 0 then, e^0 = 1 So at the yintercept or x = 0, the function becomes y = a * 1 or y = a Therefore, the constant a is the yintercept of the curve The other parameter in our equation is b If b is very small and greater than 0, the function flattens out The curve increases at a slower rate then for large b'sDrones asteroids Aerial Analysis – Challenge 1 Some humans see a photo as an image that perhaps captures a moment in time To thinking men, it can be carefully read to see what has happened in the past and perhaps what might occur in the futureOr E {\displaystyle \mathbb {E} } The expected value is also known as the expectation, mathematical expectation, mean, average, or first moment Expected value is a key concept in economics, finance, and many other subjects By definition, the expected value of a constant random variable X = c {\displaystyle X=c}
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63 Expected value If X and Y are jointly continuously random variables, then the mean of X is still given by EX = Z ∞ −∞ xfX(x)dx If we write the marginal fX(x) in terms of the joint density, then this becomes EX =S T D A U F E E L V W N X M Y Z A B K C U D B E L I E V E F Word search puzzle words to find S E E H E A R R E A D F E E L T A L K H O P E K N O W T H I N K G U E S S D R E A M B E L I E V E U N D E R S T A N D Reprinted with permission by the Dana Alliance for Brain Initiatives "FORMING WORDS" This puzzle gives you five key words toNOTES ON METRIC SPACES JUAN PABLO XANDRI 1 Introduction Let X be an arbitrary set, which could consist of vectors in Rn, functions, sequences, matrices, etc We want to endow this set with a metric;
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Answer (1 of 2) Conditional expectation is difficult to work with in the most general case Here is a link to the proof in the general case, but it may not be that informative if you are not familiar with measure theory Law of total expectation I will give you a "proof" in the special case4 the identity matrix Hence, I = C = g(t) = e(AB)te Bte At for all t After multiplying by eAteBt on both sides we have eAteBt = e(AB)t Exercises 1 If = 0, the zero matrix, prove that eA = I A 2 Use the definition (1) of the matrix exponential to prove the basic properties listed in(e) the variance of Y 4 Let Y be a random variable having mean µ and suppose that E(Y −µ)4 ≤ 2 Use this information to determine a good upper bound to P(Y −µ ≥ 10) 5 Let U and V be independent random variables, each uniformly distributed on 0,1 Set X = U V and Y = U − V Determine whether or not X and Y are
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A y″ b y′ c y = 0 Where a, b, and c are constants, a ≠ 0 A very simple instance of such type of equations is y″ − y = 0 The equation's solution is any function satisfying the equality y″ = y Obviously y1 = e t is a solution, and so is any constant multiple of it, C1 e t Not as obvious, but still easy to see, is that y 2If X and Y are independent, then E(es(XY )) = E(esXesY) = E(esX)E(esY), and we conclude that the mgf of an independent sum is the product of the individual mgf's Sometimes to stress the particular rv X, we write MTheorem 36 Let F be any partition of the set S Define a relation on S by x R y iff there is a set in F which contains both x and y Then R is an equivalence relation and the equivalence classes of R are the sets of F Pf Since F is a partition, for each x in S there is one (and only one) set of F which contains x
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Here is how you do it Given, two points, #(x_1,y_1)# and #(x_2,y_2)# and #y= ae^(bx)#X n EXjBnP(Bn) Now suppose that X and Y are discrete RV's If y is in the range of Y then Y = y is a event with nonzero probability, so we can use it as the B in the above So f(xjY = y) is de ned We can change the notation to make it look like the continuous case and writeNCSS Statistical Software NCSScom Curve Fitting – General 3515 © NCSS, LLC All Rights Reserved 7 PolyRatio(4,4) Y=(ABXCX^2DX^3EX^4) / (1FXGX^2HX^3IX^4)
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It means the slope is the same as the function value (the y value) for all points on the graph Example Let's take the example when x = 2 At this point, the y value is e 2 ≈ 739 Since the derivative of e x is e x, then the slope of the tangent line at x = 2 is also e 2 ≈ 739 x = 2 \displaystyle {x}= {2} x= 2In this setting, e 0 = 1, and e x is invertible with inverse e −x for any x in B If xy = yx, then e x y = e x e y, but this identity can fail for noncommuting x and y Some alternative definitions lead to the same function For instance, e x can be defined as → ()STAT 400 Joint Probability Distributions Fall 17 1 Let X and Y have the joint pdf f X, Y (x, y) = C x 2 y 3, 0 < x < 1, 0 < y < x, zero elsewhere a) What must the value of C be so that f X, Y (x, y) is a valid joint pdf?b) Find P (X Y < 1)c) Let 0 < a < 1 Find P (Y < a X) d) Let a > 1 Find P (Y < a X)e) Let 0 < a < 1 Find P (X Y < a)
Y A F K X D C
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= λ X∞ k=1 λ λk−1 (k −1)!Integrate 1/(cos(x)2) from 0 to 2pi;For b = 1, 2, 3, 4 Describe the graphical significance of b (b) Find the coordinates of the critical point of
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#e^b= 3/8# #b = ln(3/8)# The final equation is #y = 8e^(ln(3/8)x)# Often, the same problem is asked where the x coordinate of one of the points is not 0 When this happens, you must find the value of #b# before you find the value of #a#;EB(u)B(v) = min(u,v), we have Cov(Y(s),Y(t)) = EY(s)Y(t) −EY(s)EY(t) = EY(s)Y(t) = EsB(1/s)tB(1/t) = stEB(1/s)B(1/t) = stmin(1 s, 1 t) =min(t,s) (c) Clearly Y(t)=tB(1/t) has a continuous sample path, as B has a continuous sample path Second, as shown in part (a) the Y(t) has normal distribution with mean 0 and variance tThe fact that some of the coeffi cients are functions of x should not slow us down Applying the quadratic formula we get y = ex ± (−ex)2 − 4 1 (−ex) 2 1 ex± √ 2 4 y = 2 Our original equation is valid only for y > 0, and √ e2x 4ex > √ e2x = e
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2
Of x and t (and not y or z), so all y and zderivatives are zero yyzz0 EBEB yz y z EB00 Recall, in a chargefree medium, xxyyzz00 EBEBEB xy z x y z that is, xx00 EB xx We find and The component of the wave pointing parallel to the propagation direction does not vary along that direction Only E y and E z can vary with xD(x,y) ≥ d(y,zx)−d(x,zx) ≥ d(y,zy)−d(x,zx) = d(x,E) −d(y,E) Besides, in the case that d(x,E) ≥ d(y,E), we can get d(x,y) ≥ d(y,E) − d(x,E) So we can conclude that d(x,E) −d(y,E) ≤ d(x,y) ∀x0 ∈ X,∀ε > 0, just choose δ = ε 2, then ∀x ∈ B(x0,δ), we have d(x,E) − d(x0,E) ≤ d(xIe a way to measure distances between elements of XA distanceor metric is a function d X×X →R such that if we take two elements x,y∈Xthe number d(x,y) gives us the distance
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