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Variable X conditioned on Y = y, and consequently its expectation E(XY = y), is given Such may be the case in a twostage process, whereby the value of Y is chosen in the first stage, which then determines the distribution of X in the second stage This situation will be very commonD) E(X2e2Y) SOLUTION Using independence and the facts that EX= 1 2, E(X2) = 1 3, EY = 1 and E(Y2) = 2 gives the answers for (a)(c) as E(XY) = 3 2, E(XY) = 1 2, and E(X Y)2 = E(X2) 2EXEY E(Y2) = 4 3 For part (d) we must also compute E(e2Y) = Z 1 0 e2ye ydy= Z 1 0 eydy= 1 Therefore, E(X2e2Y) = 116 Hydrodynamics Reading #4 version 10 updated 1 ©05 A Techet 16 Hydrodynamics Prof AH Techet Potential Flow Theory "When a flow is both frictionless and irrotational, pleasant things happen"FM
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ŒC Žû"[ ȃXƒy[ƒX 100‹Ï-Eg(X) EL (X) = Ea bX = a b = L ( ) = g( ) Example 2 Bounded Random Variables Let X be a random variable with zero mean and with support on some bounded interval a;b You should convince yourself that the zero mean assumption does not matter you can always subtract the mean, ie de ne a new random variable Y = X EX and use Y inX) = 0 for all y 6= xTherefore, if we choose z =2 Ω, then ∆yΦ(y ¡
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Z) = Φ(y ¡U= @2u @x2 @2u @y2 Variable coefficients and more complex domains will be discussed in finite element methods Furthermore we assume uis smooth enough to enable us use Taylor expansion freely Given two integers m;n 2, we construct aC U(x, y) = x2/3 y1/3 Since the indifference curves are bowed towards the origin, they do obey the assumption of diminishing MRS y x 8 8 1 1 512 512 Economics 3070 d U(x, y) =min(2X, 3Y) This is an example of perfect complements The MRS is undefined at the vertex where 2X=3Y But lets graph the indifference curve, remember they L shaped
0 we have f Y(y) = 0, and for y 0 we have f Y(y) = Z 1X ~ U(0,3) exp(λ) exponential distribution f (x) = λeλx, x≥0 gamma(c, λ) gamma distribution f (x) = λ c x c1 eλx / Γ(c), x≥0 χ 2 (k) chisquare distribution f (x) = x k /21 ex/2 / ( 2 k/2 Γ(k/2) ) F (k 1, k 2) F distribution Bin(n,p) binomial distribution f (k) =We will look for the Green's function for R2In particular, we need to find a corrector function hx for each x 2 R2 , such that ∆yhx(y) = 0 y 2 R2 hx(y) = Φ(y ¡x) y 2 @R2 Fix x 2 R2We know ∆yΦ(y ¡
– Consumers Max U(x,y) subject to the budget constraint, I=Pxx P yy • Problem is made easier by the fact that we assume all variables are known with certainty – Consumers know prices and income – Know exactly the quality of the product 3 • Many cases, there is uncertainty about some variablesLet X be a –nite set of prizes , D(X) be the set of lotteries on X Let be a binary relation on D(X) Then is complete, re⁄exive, transitive and satis–es the Independence and Archimedean axioms if and only if there exists a u X !R such that, for any p,q 2D(X), p qU = x04y06 It is possible to show, and you should be able to do this, that the demand functions for x and y are px 04I x* = and py 06I y* = If income, I, is $240, you can diagram the demand curve for x A graph showing the demand curve for good x based on the utility function U = x04y06 and income of $240
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Y x z y x z y x z (A) (B) FIGURE 8 (i) x y (ii) x y (iii) x y FIGURE 9 solution The projection of curve onto the xyplane is neither a segment nor a periodic wave Hence, the correct projection is (iii), rather than the two other graphs The projection of curve (A) onto the xyplane is a vertical line, hence the corresponding projectionCIELUV is an Adams chromatic valence color space, and is an update of the CIE 1964 (U*, V*, W*) color space (CIEUVW) The differences include a slightly modified lightness scale, and a modified uniform chromaticity scale in which one of the coordinates, v′, is 15 times as large as v in its 1960 predecessorCIELUV and CIELAB were adopted simultaneously by the CIE when no clear6 0 R e s e a r c h Q u e s t i o n 7 0 C o n c l u s i o n 8 0 L i t e r a t u r e C i t e d 9 0 A p p e n d i x 2 1 0 A b s t r a c t In Los Angeles County, residents of certain communities are at risk of poorer health due to their environment Communities' relationship to health can be directly influenced by their
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Xxp yy subject to U∗= x(y 1) Findthevaluesofx and y that solve this minimization problem 4 Skippy has the following utility function u = x12 y 1 2 and faces the budget constraint M = p xxp yy (a) Suppose M =1, P y =1and P x =4 Find the optimal x and y 6To show that Xand Y are uncorrelated, we must show that Cov(X;Y) = 0, or Cov(X;Y) = EXY EXEY = EX3 EXEX2 = 0 We compute the third moment of Xusing the density function, EX3 = Z 1 1 x3p X(x) dx = Z a a x3 2a dx = (a)4 ( a)4 8a =0 Because 1=2ais constant in x, and therefore symmetric about x= 0, then every odd moment of Xwill beC ydx−xdy x2 y2 where C is a circle oriented counterclockwise (a) Show that I = 0 if C does not contain the origin Solution Let P = y x 2y 2, Q = −x x y and let D be the region bounded by C P and Q have continuous partial derivatives on an open region that contains region D By Green's Theorem, I = Z C ydx−xdy x 2y = Z C PdxQdy
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C h e y e n n e B l v d Dodson Ave One Hundredth St Reata St S a r i a A v e 85th St W 90th St W Fristad St M eyer St 117th St W Sequoia Blvd C o R d N o 6 2 9 Mc Co ne l Ave Mojave ve Ave of the Stars 78th St W Dawn Rd 2 St e c r n C o n m t h t n o Rd e r fT n L u Rd Rd e R a u u S s S Census Tract 10 CENSUS CENSUS TRACT REFERENCE MAPA(x,y) ∂u ∂x B(x,y) ∂u ∂y = C(x,y,u) (5) When A(x,y) and B(x,y) are constants, a linear change of variables can be used to convert (5) into an "ODE" In general, the method of characteristics yields a system of ODEs equivalent to (5) In principle, these ODEs can always be solved completely to give the general solution to (5)X y C 1 1 5 PROPERTIES OF LINE INTEGRALS 6 We'll use the notation Z C Mdx Ndy Parametrize the curve x= t, y= t, with tfrom 0 to 1 Put everything in terms of t dx= dt dy= dt M= x2y= t3 N= x 2y= t Now we put this into the integral I= Z C Mdx Ndy= Z 1 0 t3 dt tdt= Z 1 0 t3 tdt= 1 4
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X Y = 4 1 = 1 4 Next we redo the computation of Cov(X;Y) using the properties of covariance As usual, let X i be the result of the i th ip, so X i ˘Bernoulli(05)X √ 2 C0 √ n e−j2/n, where the sum is over all j with a √ 2n ≤ 2j ≤ b √ 2n The right hand side is the Riemann sum approximation of an integral where the intervals in the sum have length p 2/n Hence the limit is Z b a 1 C0 e−x2/2 dx This limiting distribution must be a probability distribution, so we can see that Z∞ −∞ 1Fxe u t ah cp n y s w p w q g r y n b e c s jvr u p l ns cs x t u p c s x t u p u p c p c s x t n s c n up se r a c f x e sjvr m h c h r t ns g r r c ic l s r c c s x
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IXL is the world's most popular subscriptionbased learning site for K–12 Used by over 12 million students, IXL provides personalized learning in more than 8,500 topics, covering math, language arts, science, social studies, and Spanish Interactive questions, awards, and certificates keep kids motivated as they master skillsX) for y 2Stat 110 Final Review, Fall 11 Prof Joe Blitzstein 1 General Information ThefinalwillbeonThursday12/15, from2PMto5PMNobooks, notes, computers,
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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) =Solution The closed unit ball in c 0 is not compact For example, let e k= ( nk) 1 n=1 nk= 1 if n= k 0 if n6=k245 X X 1 2 3 1 2 3 1 2 3 1 2 3 Solutionof16,#5 A A (shaded) U Solutionof17,#1 Section17 1 DrawaVenndiagramfor A (Solutionaboveright) 3 DrawaVenndiagramfor(A
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Search the world's information, including webpages, images, videos and more Google has many special features to help you find exactly what you're looking forEstimate the enthalpy of vaporization of water at 100°C from its value at 25°C (4401 kJ mol 1 ) given the constantpressure heat capacities of 7529 J K l mol l and 3358 J K l mol l for liquid and gas, respectivelyY e x Acos 2x Bsin 2x To solve for A and B using the initial values we must first differentiate y (1227) y e x Acos 2x Bsin 2x e x 2Asin 2x 2os 2x Substituting the initial values gives the equations A 2 A 2B 1, which has the solutions A 2 B 1 2 The answer thus is (1228) y e x 2cos 2x 1 2 sin 2x Case of a double root If the discriminant a2
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Let X and Y be two independent discrete random variables with the same CDFs F X and F Y Define Z = max ( X, Y), W = min ( X, Y) Find the CDFs of Z and W Solution To find the CDF of Z, we can write F Z ( z) = P ( Z ≤ z) = P ( max ( X, Y) ≤ z) = P ( ( X ≤ z) and ( Y ≤ z)) = P ( X ≤ z) P ( Y ≤ z) ( since X and Y are independentAnd the random variable Y with P(X= 100) = 1=2;Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with stepbystep explanations, just like a math tutor
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Example 5 X and Y are jointly continuous with joint pdf f(x,y) = (e−(xy) if 0 ≤ x, 0 ≤ y 0, otherwise Let Z = X/Y Find the pdf of Z The first thing we do is draw a picture of the support set (which in this case is the firstIntegrating Factors Some equations that are not exact may be multiplied by some factor, a function u (x, y), to make them exact When this function u (x, y) exists it is called an integrating factor It will make valid the following expression ∂ (uN (x, y)) ∂x =EX = is de ned as var(X) = E (X )2 The square root of the variance of a random variable is called its standard deviation, sometimes denoted by sd(X) The variance of a random variable Xis unchanged by an added constant var(XC) = var(X) for every constant C, because (XC) E(XC) = X EX, the C's cancelling It is a desirable property that
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3 y y ' 2 ' X Y FigureS13 2 Themomentgeneratingfunctionofc 1X 1 c 2X 2 is Eet(c 1X 1c 2X 2)=Eetc 1X 1Eetc 2X 2=(1−β 1c 1t) −α 1(1−β 2c 2t) −α 2 Ifβ 1c 1 =β 2c 2,thenX 1 X 2 isgammawithα=α 1 α 2 andβ=β ic i 3 M(t)=Eexp( n i=1 c iX i)= n i=1 Eexp(tc iX i)= n i=1 M i(c it) 4 ApplyProblem3withc i=1foralliThus M Y(t)= n i=1 M i(t)= n i=1 expλi(et−1Put the terms with x on one side, subtract xy on both sides `xy y' xy = 100x xy` `y' = 100x xy` Factor out x on left side `y' = x(100 y)`Z) = 0 for all y 2 Ω Now, if we choose z = z(x) appropriately, z =2 Ω, such that Φ(y ¡
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Shortcut Method for Finding the Standard Matrix Two examples 1 Let Tbe the linear transformation from above, ie, T(x 1;x 2;x 3) = 2x 1 x 2 x 3;X 1 3x 2 2x 3;3x 2 4x 3 Then the rst, second and third components of the resulting vector w, can be written respectively∂(x,y) ∂(u,v) = ∂x ∂u ∂x ∂v ∂y ∂u ∂y ∂v = 1 v − u 2 0 1 = 1 v (b) Let R be the region in the first quadrant bounded by the lines y = x, y = 2x and the hyperbolas xy = 1, xy = 2 Sketch the region S in the uvplane corresponding to R Solution The lines y = x and y = 2x in the xyplane correspond to v = u/v, v = 2u/v in
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P(X= 100) = 1=2 have the same mean EX = EY = 0 To measure the spread of a random variable X, that is how likely it is to have value of Xvery far away from the mean we introduce the variance of X, denoted by var(X) Let us consider the distance to the expected value ie, jX EXjE(XY) = E(X)E(Y) is implied by X and Y being independent, but not the other way around Question 2 You roll one red die and one green die Define the random variables X and Y as follows X = The number showing on the red die Y = The number of dice that show the number sixXyz, xy'z' are both maxterms (of 3 variables) xy' is not a maxterm because z is missing Definition (Disjunctive Normal Form) A Boolean function/expression is in Disjunctive Normal Form (DNF), also called minterm canonical form, if the function/expression is a sum
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# # E(g(X,Y))=g(x,y)f XY (x,y) It is important to note that if the function g(x,y) is only dependent on either x or y the formula above reverts to the 1dimensional case Ex Suppose X and Y have a joint pdf f XY(x,yCovariance between x and x , or y and y , or z and z would give you the variance of the x , y and z dimensions respectively Covariance n Covariance Matrix • Representing Covariance between dimensions as a matrix eg for 3 dimensions cov(x,x) cov(x,y) cov(x,z) C = cov(y,x) cov(y,y) cov(y,z) cov(z,x) cov(z,y) cov(z,z) • Diagonal is theProblem 5 Let c 0 be the Banach space of real sequences (x n) such that x n!0 as n!1with the supnorm k(x n)k= sup n2N jx njIs the closed unit ball B= f(x n) 2c 0 k(x n)k 1g compact?
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The marginal probability density function of Xis f X(x) = Z 1 1 f(x;y)dy = Z 1 jxj 1 8 (y2 yx2)e dy Z 1 jxj 1 4 ye ydy using integration by parts 1 4 jxje jx Z 1 jxj 1 4 e ydy using integration by parts 1 4 jxje jx 1 4 e jx 1 4 e jx jxj 1 Let f Y be the marginal probability density function of Y For y <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 estimators2 u(x;y) = aand lim (x;y)!(c 1;c 2) v(x;y) = b Thus the story for limits of functions of a complex variable is the same as the story for limits of real valued functions of the variables x;y However, a real variable xcan approach a real number conly from above or below (or from the left or right, depending on your point of
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BIfx, y ∈ P, then xy ∈ P c For each x ∈ R exactly one of the following holds x ∈ P, x=0, −x ∈ P The set P is the set of positive numbers Let x, y ∈ R Then x<y(read "x is less than y") if y − x ∈ P x<yis equivalent to y>x(read "y is greater than x") P = {x ∈ R x>0} xXy E(g(X,Y))g(x,y)p XY (x,y) If X and Y have a joint probability density function f XY(x,y), then !!
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