线性代数两道题,关于马可夫链和最小二乘方线,英文,求解
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线性代数两道题,关于马可夫链和最小二乘方线,英文,求解
接受中文回答,求解只为复习,希望大神帮助
接受中文回答,求解只为复习,希望大神帮助
![线性代数两道题,关于马可夫链和最小二乘方线,英文,求解](/uploads/image/z/17644288-40-8.jpg?t=%E7%BA%BF%E6%80%A7%E4%BB%A3%E6%95%B0%E4%B8%A4%E9%81%93%E9%A2%98%2C%E5%85%B3%E4%BA%8E%E9%A9%AC%E5%8F%AF%E5%A4%AB%E9%93%BE%E5%92%8C%E6%9C%80%E5%B0%8F%E4%BA%8C%E4%B9%98%E6%96%B9%E7%BA%BF%2C%E8%8B%B1%E6%96%87%2C%E6%B1%82%E8%A7%A3)
2)
xbar=(5+2+6+3)/4=4
ybar=(1+3+2)/4=1.5
β1=[∑(xiyi)-xbar* ybar/n]/[ ∑xi^2 – xbar*xbar/n]
={[5+6+0+6]-4*1.5/4} / [(25+4+36+9)-16/4]
=(17-1.5)/70=0.22
β0=ybar- β1 *xbar= 1.5-0.22*4=0.62
3)
[x y] ┌ ¼ 2/5 ┐ =[x y]
└ ¾ 3/5 ┘
x/4+(3/4)y=x
(2/5)x + (3/5)y =y
x+y=1
∴x=y=1/2
Steady state vector (1/2, ½)
The eigenvalue of T is 1, so this Markov chain converges.T^n converges.
xbar=(5+2+6+3)/4=4
ybar=(1+3+2)/4=1.5
β1=[∑(xiyi)-xbar* ybar/n]/[ ∑xi^2 – xbar*xbar/n]
={[5+6+0+6]-4*1.5/4} / [(25+4+36+9)-16/4]
=(17-1.5)/70=0.22
β0=ybar- β1 *xbar= 1.5-0.22*4=0.62
3)
[x y] ┌ ¼ 2/5 ┐ =[x y]
└ ¾ 3/5 ┘
x/4+(3/4)y=x
(2/5)x + (3/5)y =y
x+y=1
∴x=y=1/2
Steady state vector (1/2, ½)
The eigenvalue of T is 1, so this Markov chain converges.T^n converges.