HW3 Answers
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Part 1 (50 points)
A
1.
Not independent (
)
2.
Not independent (
)
3.
Independent
4.
![\begin{array}{l}
E(X)=0, E(Y)=0\\
V(X)=V(Y)=E(X^2)-E(X)^2=E(25(\cos^2T))=25E(\cos^2T)
=25\cdot 4\int_0^{\frac{\pi}{2}}\frac{1}{2\pi}\cos^2x dx\\
=\frac{50}{\pi}\int_0^{\frac{\pi}{2}}\cos^2x dx
=\left[\frac{50}{\pi}\left( \frac{1}{2}\cos x\sin x +\frac{1}{2}x \right) \right]_0^{\frac{\pi}{2}}
=\frac{50}{\pi}(\frac{\pi}{4})=12.5\\
Cov(X,Y)=0\\
\end{array}](/mediawiki/images/math/7/7/6/7761bfce7d503f8d43991a9d8226c6aa.png)
Not independent
5.
Independent
B
S1=2
S2=5
S3=1
S4=4
S5=3
Part 2 (20 points)
1
2
![\begin{array}{l}
E(X^2+Y^2)=E(X^2)+E(Y^2)=2E(X^2)=2\int_0^{10}\frac{1}{10}x^2dx \\
=2\left[ \frac{x^3}{3}\cdot \frac{1}{10}\right]_0^{10}=\frac{200}{3}=66.\dot 6\\
E(\max(X,Y))=\frac{3}{4}E(|X|)+\frac{1}{4}E(-|X|)
=\frac{3}{4}E(|X|)-\frac{1}{4}E(|X|)
=\frac{1}{2}E(|X|)=\frac{1}{2}\cdot 5=2.5\\
\end{array}](/mediawiki/images/math/7/a/9/7a9bfe04e03443cc7817070374b6c498.png)
3
![\begin{array}{l}
E(X^2+Y^2)=2E(X^2)=2(V(X)-(E(X))^2)=2V(X)=2\cdot 12.5=25\\
E(\max(X,Y))=2\int_{\frac{1}{4}\pi}^{\frac{3}{4}\pi}5\cdot\frac{1}{2\pi}\sin x dx
= \frac{5}{\pi}\left[\cos x\right]_{\frac{1}{4}\pi}^{\frac{3}{4}\pi}
=\frac{5}{\pi}\sqrt{2}=2.2508\\
\end{array}](/mediawiki/images/math/6/6/0/660e54f1d096e9e6052c110a58803537.png)
Part 3 (50 points)
Using the MATLAB normcdf function, we find that
P(N1>1) = 1 - normcdf(1,0,1) = 0.1587
P(N2>1) = 1 - normcdf(1,0,2) = 0.3085
Running mean(G>1) gives us 0.1610 . Thus, we conclude that G was created using distribution N1.
Sanov bound
p = 0.1587 q = 0.1610 m = size(G,2)
sanov = exp(exp(-m*(q*log(q/p)+(1-q)*log((1-q)/(1-p))))
1.0214e-025
Hoeffding bound
hoeff = exp(-2*(p-q)^2*m)
1.2392e-019
Chernoff bound
chernoff = exp(-m/3*(p-q)^2/p)
6.1213e-011
