Karteikarten
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Karten in diesem Stapel (3)
- 1Chapter 15 - Bivariate Normal Distributions A bivariate normal distribution (X,Y)^T=? A bivariate normal distribution has PDF? Where Q(x,y)=?(XCY)~N2[(muxCmuy),(sigma^2x psigmaxsigmayCpsigmaxsigmay sigma^2y)
fx,y(x,y) = 1/(2pisigmaxsigmaysqrt(1-p^2))*e^-[(1/2(1-p^2))]Q(x,y)
Q(x,y) = (x-mux/sigmax)^2+(y-muy/sigmay)^2-[2p(x-mux)(y-muy)]/sigmaxsigmay - 2Chapter 15 - Comments We call mu_=(muxCmuy) the? We call SIGMA=(sigma^2x psigmaxsigmayCpsigmaxsigmay sigma^2y) the? Which leads us to write (X,Y)^T~? SIGMA must be what 3 things?Mean vector
Covariance matrix
(X,Y)^T~N2(mu_,SIGMA)
1. SIGMA must be symmetric
2. SIGMA must be invertible (so SIGMA*SIGMA^-1=I2
3. SIGMA must be positive definite - 3Chapter 15 - Marginal, Conditional and Weighted Sums of Bivariate Normal Distributions Consider (X,Y)^T~N2(mu_,SIGMA). We have that: The marginal distributions are? The conditional distributions are? X and Y are independent if? aX+bY~? State the proof of the above statementsMarginal distributions are normal i.e. X~N(mux,sigma^2x) and Y~N(muy,sigma^2y)
Conditional distributions are normal i.e.
X¦Y=y~N(mux+p(sigmax/sigmay)(y-muy),sigma^2x(1-p^2))
Y¦X=x~N(muy+p(sigmay/sigmax)(x-mux),sigma^2y(1-p^2))
X and Y are independent if p=0
aX+bY~N(amux+bmuy,a^2sigma^2x+b^2sigma^2y+2abpsigmaxsigmay)
Answer on separate sheet
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- Karteikarten am Ohne Titel, erstellt von Cara Wales am 28/01/2016.
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