2017 Stats Midterm Solutions
-------------------------------

### Question 1

#1.1 33
#1.2 11
#1.3 123.5
#1.4 41

#1.5
   > 33 + 11*c(-2,2)
   [1] 11 55


### Question 2

#2.1 .6
#2.2 2.6
  > x1=14;y1=20
  > x2=26;y2=55
  > (y2-y1)/(x2-x1)
   [1] 2.916667
#2.3 -18
   > 35-2.6*20
   [1] -17
#2.4 2.5
#2.5 16.9

   > .6*2.5*11
   [1] 16.5

#2.6
   -18 + 2.6*18
   [1] 28.8


### Question 3

#3.1 .8
#3.2 
   > 1*.2+2*.5+3*.3
   [1] 2.1
#3.3
  .2*(1-2.1)^2 + .5*(2-2.1)^2 + .3*(3-2.1)^2
  [1] 0.49

   > pv = c(.2,.5,.3)
   > tv = 1:3
   > sum(pv*tv)
   [1] 2.1
   > sum(pv*(tv-2.1)^2)
   [1] 0.49

#3.4
   > sqrt(.49)
   [1] 0.7

#3.5
   > 10+5*2.1
   [1] 20.5

#3.6
   > 5^2 * .49
   [1] 12.25

#3.7
   > sqrt(12.25)
   [1] 3.5
   > 5*.7
   [1] 3.5

### Question 4

#4.1 .2
#4.2 .68
#4.3 
   > .56/(.56+.2)
   [1] 0.7368421
#4.4 No. P(Y=1) != P(Y=1 | X=1).
#4.5 X ~ Bernoulli(.76)
#4.6 .76
#4.7 
   > .76*(1-.76)
   [1] 0.1824
#4.8 No, they are not independent.
#4.9 given X=0, Y~Bernoulli(.5)
#4.10 Var(Y|X=0)=.25.
#4.11 Given X=0, var of a Bernoulli is maxed at p=.5.
#4.12 When you learn X=1 p is .74 but when you learn X=0, p=.5
so you can lean towards Y=1 when X=1 but when X=0 is is literally
a coin toss.

### Question 5

#5.1 .5
#5.2 .95
#5.3 .68
#5.4 .975
#5.5 .63
#5.6 
> 1-.63
[1] 0.37

### Question 6

#6.1 

There is no obvious pattern over time so IID Bern is plausible.

#6.2 

> .3^3
[1] 0.027

#6.3

> .7^3
[1] 0.343

#6.4
same as #6.2 .027

#6.5

The D sequences such that T=1
are
1,0,0
0,1,0
0,0,1
each has prob .3*.7^2 = 0.147
So the prob is 3*.147 = 0.441

Note, T3 is the number of 1's out of 3 ``trials''.
Tn is the number of 1's out of n trials.
If we let p be the probability of a 1,
then the number of 1's out of n trials is
called a binomial random variable.

> dbinom(3,3,.3) #chance of the 3 1's out of 3 trials, p=.3
[1] 0.027
> dbinom(1,3,.3) #change of 1 1 out of 3 trials, p=.3
[1] 0.441

e.g suppose we toss a coin 10 times,
what is the distribution of the number of heads:

> x = 0:10
> px = dbinom(x,10,.5)
> cbind(x,px)
       x           px
 [1,]  0 0.0009765625
 [2,]  1 0.0097656250
 [3,]  2 0.0439453125
 [4,]  3 0.1171875000
 [5,]  4 0.2050781250
 [6,]  5 0.2460937500
 [7,]  6 0.2050781250
 [8,]  7 0.1171875000
 [9,]  8 0.0439453125
[10,]  9 0.0097656250
[11,] 10 0.0009765625

e.g.
> .5^10
[1] 0.0009765625


### Question 7

#7.1
Assume I decide based on the expected value of my monetary
payout, if I promote I get 98 with prob .03 and -2 with prob .97.
   > .03*98 + .97*(-2) = 1
So I promote.

#7.2 
   > .03*.53
   [1] 0.0159

#7.3
   > .97*.87
   [1] 0.8439

#7.4
   > .0159+.8439
   [1] 0.8598

#7.5
   > .0159/.8598
   [1] 0.01849267
lower than the marginal prob of .03 which makes sense.

#7.6
   > 100*.0185-2
   [1] -0.15
   > .0185*98 + (1-.0185)*(-2)
   [1] -0.15

#7.7

> .03*.33
[1] 0.0099
> .97*.08
[1] 0.0776
> 0.0099/(0.0099+0.0776)
[1] 0.1131429

or just

.03*.33/(.03*.33 + .97*.08)
[1] 0.1131429

E(A) = 
0.1131429*98 + (1-0.1131429)*(-2)
[1] 9.31429
should promote.