Chris Andersen
Randomness
HW10
1)
a. x*y does not produce an eigenvector because an (n x 1) matrix (x) multiplied
by a (1 x n) matrix (y) produces an (n x n ) matrix, and in order to be a right
eigenv
1a) It is given that the variance of X and Y are 1 and because they are independent their
covariance is 0.
2c) see problem2c.m
3c) see getData.m
3d) see problem3d.m
Chris Andersen
Problem set 8
1.
a. Through induction we are able to conclude that the PMF and the expected value
are both 0.5. If you put in different values of p you will always get 0.5.
The variance
Christopher Andersen
Problemset 7
Christopher Andersen
Problemset 7
2)
a. 2x-y has values 1, 2, 4, .5, .25, .125 when plugging in
values for x = cfw_1,2,3 and y = cfw_1,2,3,4
P(1) = .1900
P(X=1,y=1) =
Christopher Andersen
Randomness
PS5
1)
a.
Christopher Andersen
Randomness
PS5
b.
2)
a. see problem2a.m
Christopher Andersen
Randomness
PS5
b. see problem2b.m
3)
a. solve (P(H | p)*P(p)/(P(H)
you can r
Christopher Andersen
Randomness
HW4
1)
a.
b.
figure 1b shows the histogram plot of the data. You see that 2 appears most often opposed
to all the values having an equal probability.
2)
Christopher And
Chris Andersen
Randomness
PS3
1)
a. S= 26^6
E = 6chose2(25^4)
P(E)= E/S
b. S=63^5
E= 53*63*63*1*1
P(E) = E/S
Chris Andersen
Randomness
PS3
c. S=8!
E = 8*2*6!
P(E) = E/S
Chris Andersen
Randomness
PS3
2
Christopher Andersen
Problem Set 2
Randomness
1)
a. In problem1a.m the script goes through 10^6 randomly generated 6-letter
words. The probability after 10^6 tries of getting the word path was around
Christopher Andersen
Problem set one
1
a) given
b) x=1-(y=z)
y=z gives you a comparison of each element of the matrix as opposed to the
entire thing like y=z would give.
2
a) check folder for tetraRol