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1. If n1 is equal to n2 (N ), the expressions for the standard deviations become:  s  1q 2    σ1 + σ22  Sp1 = N s √ (1)  2 N −1 q 2   2  √ S 1 + S2  Sp2 = N 2N − 2 The second standard deviation can be further simplified:           

s

1q 2 σ 1 + σ22 N s √ 2 1 N − 1q 2 √ √ S 1 + S22 = N 2 N −1

Sp1 = Sp2

(2)

Further simplification leads to:           

s

1q 2 σ1 + σ 22 N s 1q 2 S1 + S 22 = N

Sp1 = Sp2

(3)

So the two expressions are similar. 2. If the covariance between X1 and X2 is zero, the expressions for the slope become:  2 σ σX   2 X1 Y  b =  1  2 2 σX1 σX 2 (4) 2 σ X1 σ X2 Y    b2 =   2 σX2 1 σX 2 Further simplification leads to:

σ X1 Y σX2 σX21Y = σX2 2

    b1

=

   b2

(5)

These are the expressions for the slopes of simmple linear regressions. 3. The mean of the uniform distribution can be written as: µ=

Z x=b x=a

 x=b

b2 − a 2 x (b − a)(b + a) a+b x2  = dx = = =   2 2(b − a) b−a 2(b − a) x=a 2(b − a) (6) 1

The variance can be written as: 2

σ =

Z x=b x=a

a+b x− 2

!2

1 dx b−a

(7)

This can be rewritten as: σ2 =

Z x=b x=a

Z x=b Z x=b a+b x2 (a + b)2 dx + dx − 2x dx 2(b − a) b−a x=a x=a 4(b − a)

(8)

Solution of the integrals leads to:  x=b

(9)

  a+b b3 − a 3 (a + b)2 + (b − a) − b2 − a 2 4(b − a) 3(b − a) 2(b − a)

(10)

x=b

 x=b

a + b  (a + b)2  x3  − x2    + x σ = 4(b − a)  x=a 2(b − a) x=a 3(b − a)  x=a 2

This can be written as: σ2 =

This can be simplified as: σ2 =

(a + b)2 (b − a) (a2 + ab + b2 ) a+b + (11) − (b − a)(b + a) 2(b − a) 4 3(b − a)

Further simplification leads to: σ2 =

a2 + ab + b2 (a + b)2 (a + b)2 − + 3 2 4

(12)

a2 + ab + b2 (a + b)2 (a + b)2 − + 3 2 4

(13)

We further simplify: σ2 =

This can be rewritten as: σ2 =

4a2 + 4ab + 4b2 3a2 + 6ab + 3b2 − 12 12

(14)

We can join the two terms: σ2 =

a2 − 2ab + b2 12 2

(15)

Which leads to:

1 (b − a)2 (16) 12 We can also calculate the variance using the E[x2 ]. This can be calculated as: Z x=b x2 dx (17) E[x2 ] = x=a b − a This can be written as: σ2 =

x=b

x3  E[x ] =  3(b − a) x=a 2

This is equal to: E[x2 ] =

b3 − a 3 (b2 − a2 ) (a2 + ab + b2 ) a2 + ab + b2 = = 3(b − a) 3(b − a) 3

(18)

(19)

We can then calculate the variance: 4a2 + 4ab + 4b2 3a2 + 6ab + 3b2 a2 + ab + b2 a2 + 2ab + b2 − = − 3 4 12 12 (20) This can be simplified as: σ2 = E[x2 ]−µ2 =

σ2 =

1 a2 − 2ab + b2 = (b − a)2 12 12

(21)

4. The two main assumptions in time series analysis are stationarity (the statistical properties of the time series do not change over time) and ergodicity (calculating statistics over one realization is equal to calculating them over the entire ensemble). 5. Zt is a function of Zt−2 , so the model can be expected to conserve the autocorrelation with lag 2, as well as the variance of the process. We first derive an expression for the covariance with lag 2: γ2 = = = =

E[Zt Zt−2 ] E[(φ2 Zt−2 + at )Zt−2 ] φ2 E[Zt−2 Zt−2 ] + E[at Zt−2 ] φ2 σ 2Z 3

(22)

The second term is zero because white noise is not correlated in time. The parameter φ2 can then be written as: φ2 =

γ2 = ρ2 σZ2

(23)

We derive an expression for the variance of the process: σZ2 = = = =

E[Zt Zt ] E[Zt (φ2 Zt−2 + at )] φ2 E[Zt Zt−2 ] + E[Zt at ] φ2 γ2 + σa2

(24)

We divide this last expression by the variance of the process: 1 = φ2 ρ2 +

σa2 σ 2a 2 = ρ + 2 σZ2 σZ2

(25)

Which leads to: σa2 = (1 − ρ22 )σZ2 6.

(26)

• The mean we can simply calculate as: µX =

Z 10 0

3 X 2 (X − 10)2 dX 2500

(27)

This can be rewritten as: µX =

3 2500

Z 10 0

(X 4 − 20X 3 + 100X 2 )dX

(28)

Which is: 3 X5 20X 4 100 3 − µX = X + 2500 5 3 4 "

#10

(29)

0

Plugging in the numbers leads to: 3 100000 µX = 20000 − 50000 + 3 2500 



(30)

Or, simplified: µX =

100000  3 −30000 + 2500 3 

4

(31)

Putting everything on the same denominator: 3  −90000 + 100000  3 2500

µX = Simplified:

3 10000 3 2500 

µX =

(32)



(33)

And finally: µX = 4

(34)

• The easiest way is to calculate the variance as: i

h

2 2 σX = E X 2 − µX

(35)

We calculate the expected value of X squared as: =

Z 10

3 X 3 (X − 10)2 dX 2500

3 = 2500

Z 10

X 3 X 2 − 20X + 100 dX

(37)

 3 Z 10  5 X − 20X 4 + 100X 3 dX 2500 0

(38)

h

E X

2

i

0

(36)

Expansion leads to: h

E X

2

i

0





This can be rewritten as: h

i

E X2 =

Solution of the integral leads to: h

E X

2

i

3 X6 = − 4X 5 + 25X 4 2500 6 "

#10

(39)

0

Plugging in the numbers leads to: h

i

E X2 =

1, 000, 000 3 − 400, 000 + 250, 000 6 2500





(40)

Simplification leads to: h

i

E X2 =

 3 1, 000, 000 − 150, 000 2500 6 

5

(41)

Putting everything on the same nominator leads to: h

i

E X2 =

3 1, 000, 000 − 900, 000 2500 6

(42)

Simplification leads to: 50, 000 1 100, 0000 = 20 = 2500 2 2500 We can thus calculate the variance as: h

i

E X2 =

i

h

2 = E X 2 − µX2 = 20 − 16 = 4 σX

(43)

(44)

Thus the standard deviation is equal to 2. 7.

• Bayes’ theorem can be used to solve this problem. • Bayes: P (D|P 1)P (P1) P (D|P 1)P (P1) = P (D|P1)P (P1 ) + P (D|P2)P (P2) + P (D|P3)P (P3) P (D) (45) This results in (the students did not have to perform the actual calculation):

P(P1 |D) =

P(P1 |D) =

0.01 ∗ 0.3 = 0.158 0.01 ∗ 0.3 + 0.03 ∗ 0.2 + 0.02 ∗ 0.5

(46)

2 8. The F test statistic is SA2/SW . SA2 is the variance among the treatment means. The treatments are likely to be different if this variance is high. 2 is the variance within the treatments. The treatments are likely to SW be different if this variance is low. These variances are always positive, and the treatments are different if you divide a large number by a small number. Thus it makes sense that this test only examines the right tail.

9. The crucial property of the F-distribution is that the lower-tail critical value (FL ) is equal to 1/FU∗, with FU∗ being equal to the upper tail critical value with the degrees of freedom in the nominator and denominator swapped. In equation this becomes: FL = fα,nt −1,nt (ns −1) FU = f1−α,nt −1,nt (ns −1) ∗ F = f1−α,nt (ns −1),nt −1  U   1    FL = F U∗       

6

(47)

2 , we have nt -1 degrees of freedom If we define the test value as SA2 /SW in the nominator, and nt (ns − 1) in the denominator. If we define the test value as SA2 /S 2W , we have nt (ns − 1) degrees of freedom in the nominator, and nt − 1 in the denominator. This now becomes a left-tail test. If we calculate the critical value, this will be equal to fα,nt (ns −1),nt −1 . Equation 47 shows that this is equal to 1 divided by the lower tail critical value of the original anova definition. So the test value and the critical value are equal to 1 divided by their original formulation, so the test has to yield the same results.

10. We start by writing the regression equations:     

SXY X 2 SX SXY     X = c0 + 2 Y SY Y

= b0 +

(48)

We invert the second equation:         

SXY X SX2

Y

= b0 +

Y

S2 S2 = − Y c0 + Y X SXY SXY

(49)

The inverted slope divided by the original slope (Rs ) is: Rs =

SY2 SX2 1 = 2 2 R SXY

(50)

R is the correlation between X and Y . To compare the intercepts, we substitute the expression for the intercepts:    Y = Y − SXY X + SXY X   2 SX S 2X (51) 2 2 S S Y SXY S Y2    Y = − Y X+ Y + X  SXY SXY SY2 SXY This can be simplified as:         

Y Y

SXY SXY X+ 2 X 2 SX SX S 2Y S2 = Y − X+ Y X SXY SXY

= Y −

7

(52)

The difference between the inverted and original intercept (Di ) is: Di = Y −

SXY SY2 X −Y + 2 X SX SXY

(53)

SXY S Y2 2 − SX SXY

This can be rewritten as: Di =

!

X

(54)

S2 S2 1 − X2 Y S XY

!

(55)

Rearrangement leads to: SXY Di = 2 SX

X

This can be rewritten as: 1 Di = 1 − 2 b1 X R 



(56)

b1 is the slope of the regression of Y as a function of X. This teaches us the following: • The magnitude of the inverted slope will always be larger than the magnitude of the original slope. • If the slope of the regression and the average of the data are positive, the difference between the inverted and the original intercept will be negative. This means that the original intercept will be higher than the inverted intercept.

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