Lecture 22: Transformations and Convolutions

피망이·2024년 1월 11일
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Previous 21th review

  • Variance of Hypergeom(w, b, n)

    • p=ww+bp = \displaystyle \frac{w}{w+b}, w+b=Nw+b = N
  • Var(j=1nXj)=Var(X1)+...+Var(Xn)+2i<jCov(Xi,Xj)Var(\displaystyle \sum_{j=1}^{n} X_j) = Var(X_1) + ... + Var(X_n) + 2 \sum_{i<j} Cov(X_i, X_j)

    =np(1p)+2(n2)ww+bw1x+b1p2= np(1-p) + 2 \displaystyle {n \choose 2} \frac{w}{w+b} \frac{w-1}{x+b-1} - p^2
    =NnN1np(1p)= \displaystyle \frac{N-n}{N-1} np(1-p) ← finite population correation

    • Cov(X1,X2)=E(X1X2)(EX1)(EX2)Cov(X_1, X_2) = E(X_1 X_2) - (EX_1) (EX_2)
    • Extreme cases
      : n=1n=1Bern(p)Bern(p)의 분산
      : NN much much larger than nn, NnN11\displaystyle \frac{N-n}{N-1} \approx 1Bin(n,p)Bin(n, p)의 분산

Transformations

  • Thm.

    • Let XX be a continuous r.v. with PDF fXf_X, Y=g(X)Y = g(X)

      • gg is differentiable, strictly increasing.
    • Then the PDF of YY is given by fY(y)=fX(x)dxdyf_Y(y) = f_X(x) \displaystyle \frac{dx}{dy} where y=g(x)y = g(x), x=g1(y)x = g^{-1}(y) and this is written in terms of yy.

      • Also, dxdy=(dydx)1\displaystyle \frac{dx}{dy} = (\frac{dy}{dx})^{-1} by chain rule.
  • Proof.

    • CDF of YY is

      P(Yy)=P(g(X)y)=P(Xg1(y))=FX(g1(y))=FX(X)P(Y \le y) = P(g(X) \le y) = P(X \le g^{-1}(y)) = F_X(g^{-1}(y)) = F_X(X)

      fY(y)=fX(x)dxdy\Rightarrow f_Y(y) = f_X(x) \displaystyle \frac{dx}{dy} [chain rule]

  • Example. (log normal)

    • Y=eZY = e^{Z}. ZN(0,1)Z \sim N(0, 1)

      • dydz=ez=y\displaystyle \frac{dy}{dz} = e^{z} = y

      • fY(y)=12πe(lny)2/2f_Y(y) = \displaystyle \frac{1}{\sqrt 2 \pi} e^{-(lny)^2 / 2} ← multiply by dzdy\displaystyle \frac{dz}{dy}

      • fY(y)=1y12πe(lny)2/2f_Y(y) = \displaystyle \frac{1}{y} \frac{1}{\sqrt 2 \pi} e^{-(lny)^2 / 2}, y>0y>0 ← This is PDF.

Transformations in RnR^n

  • Y=g(X)\vec{Y} = g(\vec{X}), g:RnRng : R^n → R^n : n dimentional

    • X=(X1,...Xn)\vec{X} = (X_1, ... X_n) continuous.
  • Joint PDF of Y\vec{Y} is fY(y)=fX(x)dxdyf_{\vec{Y}}(\vec{y}) = f_{\vec{X}}(\vec{x}) |{\displaystyle \frac{d \vec{x}}{d \vec{y}}}|Jacobian, abs value of determinant.

    • dydx1=dxdy|{\displaystyle \frac{d \vec{y}}{d \vec{x}}}|^{-1} = |{\displaystyle \frac{d \vec{x}}{d \vec{y}}}|

Convolution(sums)

  • Let T=X+YT = X+Y, XX, YY indep.

    • Discrete case : P(T=t)=xP(X=x)P(Y=tx)P(T=t) = \displaystyle \sum_{x} P(X=x) P(Y=t-x)

    • Continuous case : fT(t)=fX(x)fY(tx)dxf_T(t) = \int_{-∞}^{∞} f_X(x) f_Y(t-x) dx

      • since FT(t)=P(Tt)=P(X+YtX=x)fX(x)dxF_T(t) = P(T \le t) = \int_{-∞}^{∞} P(X+Y \le t | X=x) f_X(x) dx

        =FY(tx)fX(x)dx= \int_{-∞}^{∞} F_Y(t-x) f_X(x) dx ← 미분 먼저 하고 적분하기

  • Idea : prove existence of objects with desired properties using prob.

    • Show P(A)>0P(A) > 0 for a random object.

Shannon Theory

  • Suppose each object has a "score". Show there is an object with "good" score.

    • There is an object with score is at least E(X)E(X) ← score of random object
  • Ex.

    • 100 people, 15 committees of 20, each person is on 3 committees.

      • Show there exist 2 committees with overlap 3\ge 3.
    • Idea : find average overlap of 2 random committies.

      • E(overlap)=100×(32)(152)=300(15×142)=4014=207E(overlap) = 100 \times \displaystyle \frac{3 \choose 2}{15 \choose 2} = \frac{300}{(\frac{15 \times 14}{2})} = \frac{40}{14} = \frac{20}{7}

      \Rightarrow there exists pair of committies with overlap of 207\ge \displaystyle \frac{20}{7}
      \Rightarrow have overlap of 3\ge 3


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