Quantitative analysis homework solution

 d.  P(X≤20 or X>80)

4.  X~N(100,400); i.e., X is a random variable distributed normally with its mean being equal to 100 and its standard deviation being equal to 20 (square root of 400).

 P(X≤X*)=80%. What is the value for X*? Make sure that you report the Excel command using which you computed any given probability

 P(X>X**)=60%. What is the value for X**? Make sure that you report the Excel command using which you computed any given probability

5. For the last question of your assignment, you must make sure that an internet connection is already secured, as some of the commands that you will be using in this exercise require a stable internet connection. This is an exercise in R. It also requires some peer review. And it is worth 30 points. For this exercise, you are first asked to search for the following variables in the WDI dataset (for your reference, the exact indicator names are provided in parenthesis):

You are then asked to compute the coefficient of correlation between the above variables for the US from 2001 to 2019.

a. Copy and paste the R Script that you wrote for this exercise. Include every single line, so I can also run it on my end if needed.

b. What is the coefficient of correlation between real effective exchange rate and external balance for the US from 2001 to 2019?

Peer review:

c. Compare your R Script with that of at least one of your peers. In a short paragraph (less than 200 words), describe the similarities and the differences between your codes and your peer’s codes.

d. Did your peer find the same coefficient as you?

e. For the record, including the name of your peer with whom you compared codes.

1) 

a. We know that expected value is the sum of product of X and their probabilities. Thus, Expectation = ∑X*P(X)=0*0.1+0.5*0.3+1*0.5+5*0.1=0.15+0.5+0.5=1.15

Interpretation: Here X is pounds; thus we can say that on average X is going to be equal to the expected value i.e. 1.15 Pounds.

b. Variance is the squared deviation of random variable i.e. Variance = E(X^2 )-E(X)^2. We know E(X) = 1.15, we’ll be calculating E(X^2 ) to find Var(X). So,

E(X^2 )=∑X^2*P(X)=0*0.1+0.25*0.3+1*0.5+25*0.1=0.075+0.5+2.5=3.075

Thus, Var(X) = E(X^2 )-E(X)^2= 3.075- 〖1.15〗^2= 3.075-1.3225=1.7525

c. Standard deviation is the square root of variance. Thus,

σ_x=√(var(X) )=√1.7525=1.3238.

Interpretation: Standard deviation measures how spread X is. Here the standard deviation is 1.3238 and the expected values is 1.15, then we can say that

 approximately 68% of the data lies 1 standard deviation from the mean i.e.,(-0.174, 2.474)

 approximately 95% of the data lies 2 standard deviations from the mean i.e.,in the interval (-1.498, 3.798)

 approximately 99.7% of the data lies 3 standard deviations from the mean i.e.,in the interval (-2.822, 5.122)

d. Units of measurement i) Expected value = pounds

ii) Variance = pounds^2as squared deviations are calculated.

iii) Standard deviation = Pounds as it’s square root of variance

2)

a. The joint probability distribution shows that there’s a only 4 combinations of (X,Y) which has positive probability of occurrence. Thus, only four combinations of X and Y are possible. Both X and Y starts from 0 and increase together. Though the increase in Y is much more than X, but they both show an increasing pattern together. Thus, we can say that there is a positive correlation between the two variables.

b. We need to calculate E(Y) and then will use it in the Covariance formula.

E(Y)= ∑Y*P(Y)=0*0.1+100.55*0.3+195.75*0.5+400*0.1= 168.04

Now, we will compute the correlation using

ρ_XY=∑_(i=1)^N▒〖(X_i-E(X) )×(Y_i-E(Y) )×P(X=X_i,Y=Y_i ) 〗=(0-1.15)*(0-168.04)*0.1+(0.5-1.15)*(100.55-168.04)*0.3+(1-1.15)*(195.75-168.04)*0.5+(5-1.15)*(400-168.04)*0.1

=19.3246+13.165055-2.07825+89.3046 =119.7115

c. The unit of measurement is pounds*$. It’s not making any sense, thus, it’s meaningful.

d. As we know Var(Y) = E(Y^2 )-E(Y)^2. We know E(Y) = 168.04. thus, we will calculate E(Y)^2.

E(Y^2 )= ∑Y^2*P(Y)=0*0.1+〖100.55〗^2*0.3+〖195.75〗^2*0.5+〖400〗^2*0.1=3033.091+19159.03+16000=38192.12

Thus, var(Y) = E(Y^2 )-E(Y)^2=38192.12-〖168.04〗^2=9954.678

Hence the standard deviation is: σ_y=√(var(Y) )=√9954.678=99.773. Now, we can calculate the correlation:

r_XY=ρ_XY/(σ_X×σ_Y )=119.115/(1.324*99.773)=0.902

e. There is no unit of measurement of the correlation as both numerator and denominator have the same unit of measurement i.e. pounds*$. The unit of measurement is meaningful, as the correlation tells us the direction of linear relationship between the variables.

f. The obtained correlation coefficient is 0.902. Using this we can say that there is a strong positive correlation between the two variables pounds and dollar. Hence, we can say that on average as X increases or decrease,we can expect Y also to increase or decrease. In answer of part (a) we expected a positive correlation between the variables, and here we obtained a correlation coefficient of 0.902 which is positive. Thus, the result is obtained as expected.

3) Here the bold italic fonts represent the Excel command, while the non-italicized are the answers:

a. P((X≤75) = NORM.DIST(75,100,20,TRUE) = 0.10565