Sample Cheat sheet for AQMF                                                              represented by N(µ,σ2). “Bell shaped” curve that is symmetric about the
                                                                                                 mean (mean, median and mode are all the same). within ± 1/2/3 standard
        S-Sample space, E an event in, or subset of S, Pr(E) = n(E). If Pr(E) = 1                deviations of the mean there are 68%/ 95%/ 99.7% of the values. Total area
                                                                       n(S)                      under the curve is 1 and area between x and x (x ≤ x ) is Pr(x ≤ X ≤ x ).
        then E is certain to occur, if Pr(E) = 0 then E is impossible. Intersection                                                       1       2   1     2         1          2
                                                                                                 The standard normal, N(0,1), has µ = 0 and σ = 1. Given a value x from
        of E and F, E ∩ F = {x ∈ S|x ∈ E and x ∈ F}. Union of E and F,                           N(µ,σ2), P(X ≤ x) = P(Z ≤ z) where z = x−µ is a value from N(0,1)
        E∪F ={x∈S|x∈E orx∈F}. ThecomplementofE,Ec ={x∈S|x6∈E}.                                                                                         σ
        E∩Ec=φ(theemptyset)andE∪Ec =S. SoPr(Ec)=1−Pr(E). Inclusion-                              and P(Z ≤ z) is obtainable from a table or calculator. Binomial distribution
                                                                                                 B(n,p) is approximately normal if p is not close to 0 or 1 and n large enough,
        exclusion principle Pr(E∪F) = Pr(E)+Pr(F)−Pr(E∩F). Events E and F                                                                          p
        are mutually exclusive if E ∩ F = φ. prob of A given B: Pr(A|B) = Pr(A∩B)                with mean µ = np and standard deviation σ =          np(1−p).
                                                                                    Pr(B)        Two-Variable Calculus z = f(x,y)
        or Pr(A∩B)=Pr(A|B)Pr(B)=Pr(B|A)Pr(A). prob trees, the prob of one                        Contour curve- the intersection of horizontal plane z = k and z = f(x,y).
        path is the product of probabilities on all of the branches along the path. If an        Level curve- projection of a contour curve onto the xy-plane. Partial deriva-
        event can be described by more than one path, then the prob of the event is              tive of f with respect to (w.r.t.) x, f (x,y): treat y as constant and differentiate
        the sum of the prob for each path. Bayes formula Pr(B|A) = Pr(A|B)Pr(B).                                                      x
                                                                                 Pr(A)           w.r.t. x. f (x,y) = lim       f(x+h,y)−f(x,y) = ∂z = ∂f. f (x,y) is the slope of
        Events A and B are independent if Pr(A|B) = Pr(A) and Pr(B|A) = Pr(B).                              x             h→0        h           ∂x    ∂x    x
                                                                                                 f at (x,y) in the direction of x. The partial derivative of f w.r.t. y, fy(x,y),
        Events B ,B ,...B are: exhaustive if         B ∪B ∪...∪B =S,                                                                                                             2
                  1   2      n                         1    2           n                        is defined similarly. Second order partial derivatives: f (x,y) = ∂ f,
        disjoint if  B ∩B =φifi6=j                                                                                                                                 xx           ∂x2
                       i    j                                                                    f (x,y) = ∂2f. In general, f      = ∂2f =f = ∂2f this can be used to check
         and if both, thenPr(A) = Pr(A∩B )+Pr(A∩B )+···+Pr(A∩B )                                  yy          ∂y2               xy   ∂x∂y     yx    ∂y∂x
                                                1              2                    n            calculations.
        Markov chains: p         the prob of moving from state i to state j at the next
                              i,j                                                                Critical points (a,b) are where both f (a,b) = 0 and f (a,b) = 0.             Sec-
        time step. The matrix P = [p ] is called the transition matrix. For each                                                            x                   y
                                          i,j                                                    ond derivative test, if (a,b) is a a critical point ∆(a,b) = f    (a,b)f (a,b) −
        row of P, summing elements in a row gives 1. If Xn is a prob vector then                                                                               xx       yy
                                       n                                                                      f  (a,b)f   (a,b) 
        X =X P orX =XP . Factorial n denoted by n! = 1.2.3...n. The                                2           xx       xy      
           n      n−1        n      0                                                            f (a,b) =
                             n
       n!                                                         xy          f  (a,b)f (a,b) 
        binomial coefficient k = k!(n−k) is the number of ways you can select k items                             xy       yy       (a,b)
        out of n distinct items. Prob of k sucesses from n trials each with prob of              f has a relative min at (a,b) if ∆(a,b) > 0 and f      (a,b) > 0 relative max at
                                    
                                                                                                                xx
                                    n   k       n−k                                              (a,b) if ∆(a,b) > 0 and f    (a,b) < 0 and Saddle pt. if ∆(a,b) < 0, test incon-
        success p is P(X = k) = k p (1−p)           .                                                                       xx                                  df    ∂f dx   ∂f dy
          Summary statistics: Given a population of N values and sample from                     clusive if ∆(a,b) = 0.   chain rule for multivariable fns dt = ∂x dt + ∂y dt
                                                                                  P
        the population of n < N values x ,...,x , sample mean: x¯ = 1                n  x ,      Lagrange Multipliers for constrained optimisation. Optimise f(x,y) with
                                               1       n                        n    i=1 i
                                      P                                           P
                            2      1    n           2                           1    N           constraint g(x,y) = 0 form Lagrangian L(x,y,λ) = f(x,y)+λg(x,y) then find
        sample variance: s =               (x −x¯) , population mean: µ =               x ,
                            x    n−1    i=1   i                                 N    i=1 i                         ∂L       ∂L           ∂L
                                        P                                                        (a,b) such that      =0,      =0and         =0this gives all possible points for
                                 2    1    N          2                                                            ∂x       ∂y           ∂λ
        population variance: σ =              (x −µ) .     Population standard deviation
             √                        N    i=1  i                                                maxima and minima.
        σ =    σ2.
        A discrete random variable X has expected value E(X) = Px Pr(x ) and                     Calculus                                  f(b)−f(a)
                                     P                                        i     i            Average fn value between a and b is                instantaneous rate of change
                                2              2                                                                                             b−a
        variance var(X) = σ =          (x − µ) Pr(x ). Mode- most frequent does not
                                                      i                                          is called the derivative.     The derivative from first principles is f′(x) =
        have to be unique. Median given data ordered from smallest to largest, the               lim     f(x+h)−f(x). Marginal revenue is the derivative of revenue w.r.t q, sim-
        median is the value that has half of the data below and half the data above                  h→0      h
        it.  If this point falls between two data points, the median is the average              ilarly cost and profit. The derivative is also the slope of the tangent at a
        of the two values. Normal distribution: with mean µ and variance σ2 is                   point. Second derivative, the result of differentiating a fn twice. Implicit dif-
                                                                                             1
        ferentiation: treat y as a fn of x and use chain rule to differentiate leaving            then find the solution by R(1/g(y)).dy = R f(x).dx. Matlab int command:
        the derivative of y as dy. f is increasing if x < x then f(x ) < f(x ) or                Rb
                                 dx                       1     2           1         2           a f(x).dx syms x; int(f(x),x,a,b) or for indefinite integral just syms x;
        decreasing then f(x ) > f(x ). If f′(x) > 0 then f is increasing, f′(x) < 0
                              1        2                                                         int(f(x),x).
        then f is decreasing. Critical point x = c occurs when f′(c) = 0 or f′(c) is             Riemann sums, used to approximate integrals like upper and lower sums.
        undefined. A critical point is a local max if f′(x) > 0 for x < c and f′(x) < 0           Some examples are Left hand end pt.LE = ∆xPf(x                 ), Right hand end
                                                                                                                                             n               i−1
        for x > c, conversley for minimum. Second derivative test, min if f′′(c) > 0             pt. RE = ∆xPf(x) and mid-point sum M = ∆xPf((x +x                            )/2) .
        and max if f′′(c) < 0. when f′′(x) = 0 test is inconclusive.                                      n              i                          n                i    i−1
                                                                                                 Trapezoidal rule T = (LE +RE )/2. Error in approximation for mid-point
                                  ′′                          ′′                                                     n        n       n
        Pt. of inflection when f (x) = 0. Concavity: if f (x) > 0 f is concave up, or             and trapezoidal is roughly 1/n2
        f′′(x) < 0, concave down. If (c,f(c)) is a point of inflextion and f goes from
        concave up to concave down then c is the point of diminishing returns. f(d) is                               index laws                      log laws
                                                                                                                n
        the absolute max if f(d) ≥ f(x) ∀x, similarly for absolute min.                                        a =a.a...a n times          log(AB) = log(A)+log(B)
                                                                                                                    n m      n+m
        Profit = revenue - cost. Average cost = total cost divide quantity c(x)/x                                   a a =a                 log(A/B) = log(A)−log(B)
        Upper and Lower Sum, partition the interval (a,b) into n equal parts with                                   n  m      n−m                     n
                                                                                                                  a /a =a                        logA =nlogA
        interval width ∆x = b−a so the ith interval is [x       , x ] where x = a + i∆x                                 n    n   n                      lnb    logc b
                                 n                           i−1   i          i                                   (a/b) = a /b                 log b =      =
        M maximum on the ith interval and m the minimum on the ith interval.                                                 √                    a     lna    logc a
           i            P                  P         i                                                        −n       n     n       1/n
        So U = ∆x         M, L = ∆x m if A is the area under the graph be-                                  a    =1/a          a = a                 log1 = 0
              n              i   n              i
                                                                          U +L
        tween a and b then L ≤ A ≤ U . An Estimate for A is                n   n, with er-                   P            P      P                    P
                                 n           n                               2                                  a +b =      a +     b                   n   1 = n
                             Un−Ln                                                                               i    i       i      i                  i=1
        ror in estimate =           . If f is increasing on (a,b) then M = f(x ) and                            P            P                   P
                                2            R                              i       i                             (ka ) = k(    a )                 n  i = n(n+1)/2
                                              b                                                                       i          i                  i=1
        m =f(x ). Definite integral              f(x).dx = A the area under f between a                   P          P             P         P
           i      i−1                         a                                                             k          n            n          n   2
                                  R                                     R                                      b +           b =        b         i  =n(n+1)(2n+1)/6
        and b. If f(x) < 0 then     b f(x).dx < 0 Area between fns is    (f(x)−g(x)).dx                     i=1 i      i=k+1 i      i=1 i      i=1
                                   a                                          ′
        where f(x) > g(x). Fundamental theorem of Calculus if F (x) = f(x)                                         integral of f           f              df/dx
        then Rbf(x).dx = F(b) − F(a) Average Fn value between (a,b) is given by
             R a                                                                                                        cx                 c                0
          1   b f(x).dx                                                                                             n+1               n                     n−1
         b−a a                                                                                                     x    /(n+1)      x for n 6= −1        nx
          ∗  ∗                                                                                                                                                 2
        p ,q denote equilibirum price and qantity. Consumer surplus- the total                                          1/x              ln|x|           −1/x
                                                                       Rq∗           ∗                                    x                 x                x
        amount saved by consumers buying at equilibrium price = 0 (D(q)−p ).dq.                                         e                  e               e
        Producer surplus, the benifit the producer gets by selling at equilibrium                                       x                    x              x
                  R ∗                                                                                                 a /lna               a             a lna
        prices =    q (p∗ −S(q)).dq                                                                                  kR f(x)             f(x)            kf′(x)
                   0                                                                                             R         R
        Present and future value, P present value, S future value, t years interest rate                           f(x)+ g(x)         f(x)+g(x)       f′(x) +g′(x)
                                                   −rt            rt
        r compounded continuously so P = Se            or S = Pe . Generalising if f(t)                                                                √
        represents a continous income stream in dollars for annum of k years, Present                                  2                                   2
                              Rk       −rt                            kr R k    −rt              Quadratic law: if ax +bx+c = 0 then x = (−b±             b −4ac)/2a
        value of f(t) is P = 0 f(t)e      .dt and future value S = e     0 f(t)e    .dt          product rule: d/dx(f(x)g(x)) = f′(x)g(x)+f(x)g′(x)
        Differential Equations (DE) an equation in which the Derivative of an un-                                                        ′                ′          2
                                                                                                 quotient rule: d/dx(f(x)/g(x)) = (f (x)g(x)−f(x)g (x))/(g(x) )
        known function occurs. Ordinary DE is a DE in which the unknown Fn is                    integration by parts: R uv′.dx = uv − R u′v.dx
        of 1 variable, and the order of a DE is the highest derivative that occurs. so                                                                                  
        if the DE is f′(x) = kf(x) then f(x) = Aekx. Seperable DE dy = f(x)g(y)                                 a b                                       1        d   −b
                                                                            dx                          A= c d             det(A) = ad−bc      A−1 = ad−bc        −c    a
                                                                                             2