Portfolio Management - SSEI · 2019-08-07 · An investor is considering the following two alternatives for investing in the above stocks: i. Place equal proportion of his money in

Strategic FinancialManagement

Portfolio Management

Extra Practice Questions

Strategic Financial Management

1SANJAY SARAF SIR

PROBLEM - 1

Presently firms has debt - equity ratio of 2 and a & a of 2.2. The firm now decides tochange the debt-equity ratio to a level of 1.2. What would be its new use taxrate = 30%.

Solution :

Step 1: Deleveraging

1 1

uu D t

E

2 2

1 2 0 7

..

2 22 4U..

0 92 U .

Step 2 : Re leveragary

1 1

L uDB B tE

0 92 1 1 2 0 7 . . .

L 0.92 1 0.84β L 1.69β

PROBLEM - 2realRF 3%

Inflation premium = 6.5%mR 15%

i. What is the equation of the SML.ii. What is the new equation of the SML if inflation goes up to 7.5%iii. What is the equation of the SML if market risk premium goes up by 20% of its

current level.iv. What if there is a combination of ii & iii


1SANJAY SARAF SIR

PROBLEM - 1


Solution :


1 1

uu D t

E

2 2

1 2 0 7

..

2 22 4U..

0 92 U .


1 1

L uDB B tE

0 92 1 1 2 0 7 . . .

L 0.92 1 0.84β L 1.69β






1SANJAY SARAF SIR

PROBLEM - 1


Solution :


1 1

uu D t

E

2 2

1 2 0 7

..

2 22 4U..

0 92 U .


1 1

L uDB B tE

0 92 1 1 2 0 7 . . .

L 0.92 1 0.84β L 1.69β





Portfolio Management - Additional Practice Questions

2SANJAY SARAF SIR

Solution :

i. We have fR nominal = realR inf lation= 3 + 6.5= 9.5%

mR = 15% (it is important to know that this is also nominal i.e it includes 6.5%)

Market Risk Premium m fR R= 15 - 9.5 = 5.5%

SML e f m fR R R R β eR 9.5 5.5β

ii. Now, inflation premium = 7.5% f no min alR 3 7.5%

= 10.5%

mR = 15 + 1 = 16%

m fR R = 16 - 10.5= 5.5% (Same as earlier)

New SML= eR = 10.5 + 5.5

iii. New m fR R = 5.5 + 20% of 5.5= 6.6%

SML = eR 9.5 5.5β

iv. eR 10.5 6.6β

Note : The whole purpose of the above e.g. was to tell you that when inflationchanges, m fR R will not change.


2SANJAY SARAF SIR

Solution :






= 10.5%

mR = 15 + 1 = 16%


New SML= eR = 10.5 + 5.5

iii. New m fR R = 5.5 + 20% of 5.5= 6.6%

SML = eR 9.5 5.5β

iv. eR 10.5 6.6β



2SANJAY SARAF SIR

Solution :






= 10.5%

mR = 15 + 1 = 16%


New SML= eR = 10.5 + 5.5

iii. New m fR R = 5.5 + 20% of 5.5= 6.6%

SML = eR 9.5 5.5β

iv. eR 10.5 6.6β



3SANJAY SARAF SIR

PROBLEM - 3

Investible fund = 5,00,000 on 1st Jan, 2017.Client has mentioned that portfolio should not fall by more than 20%. Client hasagreed to a multiplier of 1.5. Rebalancing is to be done every quarter. 8fR %perannum compound quarterly.Current share price = 300.Share price on 1st April = 330.Share price on 1st July = 360.

Calculate the allocation of the portfolio into equity & bond initially & on eachrebalancing date.

Solution :

I. 1st Jan

A = 500000F = 80% 5,00,000

= 4,00,000M = 1.5E = m(A - F)

= 1.5(5,00,000 - 4,00,000) 1,50,000}Bond = 350,000

II. 1st April = New Price, 330.

Equity = 330150000 165000300

Bond = 3,50,000 1.02 = 357000A = E + B 5,22,000Floor 4,00,000 1.02

408000A - F = 114000 Equity = 1.5 114000

= 1,71,000


3SANJAY SARAF SIR

PROBLEM - 3



Solution :

I. 1st Jan

A = 500000F = 80% 5,00,000

= 4,00,000M = 1.5E = m(A - F)

= 1.5(5,00,000 - 4,00,000) 1,50,000}Bond = 350,000


Equity = 330150000 165000300

Bond = 3,50,000 1.02 = 357000A = E + B 5,22,000Floor 4,00,000 1.02

408000A - F = 114000 Equity = 1.5 114000

= 1,71,000


3SANJAY SARAF SIR

PROBLEM - 3



Solution :

I. 1st Jan

A = 500000F = 80% 5,00,000

= 4,00,000M = 1.5E = m(A - F)

= 1.5(5,00,000 - 4,00,000) 1,50,000}Bond = 350,000


Equity = 330150000 165000300

Bond = 3,50,000 1.02 = 357000A = E + B 5,22,000Floor 4,00,000 1.02

408000A - F = 114000 Equity = 1.5 114000

= 1,71,000


4SANJAY SARAF SIR

Bond = 522000 - 1710000 = 351000 Action Transfer `6000 from Bond to Equity

III. 1st July = New Price = 360

Equity = 360171000 186545330

Bond = 351000 1.02 = 358020A = E + B 544565Floor = 408000 1.02 = 416160A - F 128405Equity = 1.5 128405

= 192608Bond = 544565 - 192608

= 351957Action :- Transfer ` 6063 from Bond to Equity.

PROBLEM - 4

A conservative investor is analyzing the shares of PSEL which is currently trading atRs. 1,180. For the year 1999 - 2000, the earnings per share (EPS) was Rs. 40. Theinvestor has generated the following scenarios for the next year with thecorresponding probabilities:

P/E ratio EPS 20 305060

0.200.30

0.350.15

You are required to calculate the expected risk and return for the share of PSPEL


4SANJAY SARAF SIR



Equity = 360171000 186545330


= 192608Bond = 544565 - 192608


PROBLEM - 4



0.200.30

0.350.15



4SANJAY SARAF SIR



Equity = 360171000 186545330


= 192608Bond = 544565 - 192608


PROBLEM - 4



0.200.30

0.350.15



5SANJAY SARAF SIR

Solution :

The current price of PSEL is Rs. 1180. The EPS is Rs. 40. Hence, the P/E ratio is

1180 29.5

40

The various EPS and P/E ratios are given below.

(1)EPS

(2)P/E Ratio

(3)Probability

(4)Expected

Price (1 × 2)

(5)Expected

Return (Rs)

(6)Expected

Return (%)50 20 0.20 1000 -180 (15.25)50 30 0.35 1500 320 27.1260 20 0.30 1200 20 1.6960 30 0.15 1800 620 52.24

The expected return will be

E (P) = (-15.25 × 0.2) + (27.12 × 0.35) + (1.69 × 0.30) + (52.54 × 0.15)= -3.05 + 9.49 + 0.507 + 7.88= 14.827 or 14.83%

The risk of the stock is found below.

X X - X 2X - X Prob 2X - X ×Prob

(15.25) (30.08) 904.81 0.20 180.9627.12 12.29 151.04 0.35 52.86

1.69 (13.14) 172.66 0.30 51.8052.54 37.71 1422.04 0.15 213.31

498.93

22p 498.93 %σ


5SANJAY SARAF SIR

Solution :


1180 29.5

40


(1)EPS

(2)P/E Ratio

(3)Probability

(4)Expected

Price (1 × 2)

(5)Expected

Return (Rs)

(6)Expected

Return (%)50 20 0.20 1000 -180 (15.25)50 30 0.35 1500 320 27.1260 20 0.30 1200 20 1.6960 30 0.15 1800 620 52.24


E (P) = (-15.25 × 0.2) + (27.12 × 0.35) + (1.69 × 0.30) + (52.54 × 0.15)= -3.05 + 9.49 + 0.507 + 7.88= 14.827 or 14.83%



(15.25) (30.08) 904.81 0.20 180.9627.12 12.29 151.04 0.35 52.86

1.69 (13.14) 172.66 0.30 51.8052.54 37.71 1422.04 0.15 213.31

498.93

22p 498.93 %σ


5SANJAY SARAF SIR

Solution :


1180 29.5

40


(1)EPS

(2)P/E Ratio

(3)Probability

(4)Expected

Price (1 × 2)

(5)Expected

Return (Rs)

(6)Expected

Return (%)50 20 0.20 1000 -180 (15.25)50 30 0.35 1500 320 27.1260 20 0.30 1200 20 1.6960 30 0.15 1800 620 52.24


E (P) = (-15.25 × 0.2) + (27.12 × 0.35) + (1.69 × 0.30) + (52.54 × 0.15)= -3.05 + 9.49 + 0.507 + 7.88= 14.827 or 14.83%



(15.25) (30.08) 904.81 0.20 180.9627.12 12.29 151.04 0.35 52.86

1.69 (13.14) 172.66 0.30 51.8052.54 37.71 1422.04 0.15 213.31

498.93

22p 498.93 %σ


6SANJAY SARAF SIR

PROBLEM - 5

Consider the following information relating to the returns from two stocks and themarket index in different economic scenarios:

Scenario Probabilityof scenario

Stock A (%) Stock B (%) Return frommarket index (%)

BoomSlowgrowthStagnation

0.250.100.450.20

-15193515

-8-52518

-7122025

From the above information, you are required to :

a. Calculate the ex-ante beta for the two stocks

b. Assuming that SML holds good, determine the Alpha of the two stocks andcomment on the same.

Also assume a risk free rate of interest of 7%.

Solution :

a. Market

RM Pi RMPi RM-E(RM) [RM-E(RM)]2

Square of deviations×Pi

-0.070.120.200.25

0.250.100.450.20

–0.01750.01200.09000.0500

–0.2045–0.01450.06550.1155

0.04180.00020.00430.0133

0.010500.000020.001900.00270

0.1345 0.01512

Expected Return on market = 13.45VarM = 0.01512OM = 12.30%


6SANJAY SARAF SIR

PROBLEM - 5





0.250.100.450.20

-15193515

-8-52518

-7122025





Solution :

a. Market



-0.070.120.200.25

0.250.100.450.20

–0.01750.01200.09000.0500

–0.2045–0.01450.06550.1155

0.04180.00020.00430.0133

0.010500.000020.001900.00270

0.1345 0.01512



6SANJAY SARAF SIR

PROBLEM - 5





0.250.100.450.20

-15193515

-8-52518

-7122025





Solution :

a. Market



-0.070.120.200.25

0.250.100.450.20

–0.01750.01200.09000.0500

–0.2045–0.01450.06550.1155

0.04180.00020.00430.0133

0.010500.000020.001900.00270

0.1345 0.01512



7SANJAY SARAF SIR

Stock A

RA

(1)Pi

(2)RAPi

(3)RA-E(RA)

(4)RM-E(RM)

(5)Product

(6) = (4) × (5)Product × Pi

(7) = (6)× (2)-0.150.190.350.15

0.250.100.450.20

–0.03750.01900.15750.0300

–0.3190.021

0.181 –0.019

–0.2045–0.01450.06550.1155

0.0652–0.00030.0119

–0.0022

0.0163–0.00003

0.0054–0.00043

0.1690 0.02124

Expected return on stock A = A iR P 16.9%

Stock B

BR(1)

iP(2)

B iR P(3)

B BR E R(4)

M MR E R(5)

Product(6) = (4) × (5)

Product x Pi

(7) = (6) × (2)–0.08–0.050.250.18

0.250.100.450.20

–0.0200–0.00500.11250.0360

–0.2035–0.17350.12650.0565

–0.2045–0.0145+0.06550.1155

0.04160.00250.00830.0065

0.010400.000250.003700.00130

0.1235 0.01565

B iExpected Return on Stock B = R P = 12.35%

A A M M iAMA

M M

R - E R R - E R PCov 0.02124Beta = = = =1.40Var Var 0.01512

BMB

M

Cov 0.01565Beta = = =1.04Var 0.01512

b. A f A M fR = R +β R -R

= 7 + 1.4 (13.45 – 7) = 16.03

A A=E R -Required return

= 16.9 – 16.03 = 0.87

As alpha is positive, Stock A is under valued

BR =7 + 1.04 (13.45 - 7) = 13.71

B =12.35 - 13.71 = - 1.36

As alpha is negative, Stock B is overvalued.


7SANJAY SARAF SIR

Stock A

RA

(1)Pi

(2)RAPi

(3)RA-E(RA)

(4)RM-E(RM)

(5)Product

(6) = (4) × (5)Product × Pi

(7) = (6)× (2)-0.150.190.350.15

0.250.100.450.20

–0.03750.01900.15750.0300

–0.3190.021

0.181 –0.019

–0.2045–0.01450.06550.1155

0.0652–0.00030.0119

–0.0022

0.0163–0.00003

0.0054–0.00043

0.1690 0.02124


Stock B

BR(1)

iP(2)

B iR P(3)

B BR E R(4)

M MR E R(5)

Product(6) = (4) × (5)

Product x Pi

(7) = (6) × (2)–0.08–0.050.250.18

0.250.100.450.20

–0.0200–0.00500.11250.0360

–0.2035–0.17350.12650.0565

–0.2045–0.0145+0.06550.1155

0.04160.00250.00830.0065

0.010400.000250.003700.00130

0.1235 0.01565


A A M M iAMA

M M


BMB

M

Cov 0.01565Beta = = =1.04Var 0.01512


= 7 + 1.4 (13.45 – 7) = 16.03


= 16.9 – 16.03 = 0.87


BR =7 + 1.04 (13.45 - 7) = 13.71

B =12.35 - 13.71 = - 1.36



7SANJAY SARAF SIR

Stock A

RA

(1)Pi

(2)RAPi

(3)RA-E(RA)

(4)RM-E(RM)

(5)Product

(6) = (4) × (5)Product × Pi

(7) = (6)× (2)-0.150.190.350.15

0.250.100.450.20

–0.03750.01900.15750.0300

–0.3190.021

0.181 –0.019

–0.2045–0.01450.06550.1155

0.0652–0.00030.0119

–0.0022

0.0163–0.00003

0.0054–0.00043

0.1690 0.02124


Stock B

BR(1)

iP(2)

B iR P(3)

B BR E R(4)

M MR E R(5)

Product(6) = (4) × (5)

Product x Pi

(7) = (6) × (2)–0.08–0.050.250.18

0.250.100.450.20

–0.0200–0.00500.11250.0360

–0.2035–0.17350.12650.0565

–0.2045–0.0145+0.06550.1155

0.04160.00250.00830.0065

0.010400.000250.003700.00130

0.1235 0.01565


A A M M iAMA

M M


BMB

M

Cov 0.01565Beta = = =1.04Var 0.01512


= 7 + 1.4 (13.45 – 7) = 16.03


= 16.9 – 16.03 = 0.87


BR =7 + 1.04 (13.45 - 7) = 13.71

B =12.35 - 13.71 = - 1.36



8SANJAY SARAF SIR

PROBLEM - 6Suppose the assumptions of CAPM are valid and unlimited borrowing and lending atrisk-less rate of interest is possible. You are required to determine the unknownquantities in the following table.

Stock ExpectedReturn (%)

StandardDeviation (%)

Beta Unsystematic Risk(%)2

Super Cements 12 ? 1.52 15Cresent Pharma ? 8 0.96 9DFL Plastics 9 ? 0.80 25

Solution :

According to CAPM

iR = f m fR β R R

AR = f A m fR β R R ..............................(I)

cR = f c m fR β R R ...............................(II)

(I) - (II)

A CR R = A C m fβ β R R

4 = m f1.75 0.60 R R

4 = m f1.15 R R

41.15

= m f3.478 R R

Now putting the value of m fR R in equation (I)

14 = fR 1.75 3.478

fR = 14 6.087 7.913%

BR = 7.913 0.82 3.478 = 10.76%

Total risk = Systematic risk + Unsystematic risk2βσ = 2 2 2

B m cBβ σ +σ

26 = 2 2B mβ σ +5


8SANJAY SARAF SIR






Solution :

According to CAPM

iR = f m fR β R R

AR = f A m fR β R R ..............................(I)

cR = f c m fR β R R ...............................(II)

(I) - (II)


4 = m f1.75 0.60 R R

4 = m f1.15 R R

41.15

= m f3.478 R R


14 = fR 1.75 3.478

fR = 14 6.087 7.913%

BR = 7.913 0.82 3.478 = 10.76%


B m cBβ σ +σ

26 = 2 2B mβ σ +5


8SANJAY SARAF SIR






Solution :

According to CAPM

iR = f m fR β R R

AR = f A m fR β R R ..............................(I)

cR = f c m fR β R R ...............................(II)

(I) - (II)


4 = m f1.75 0.60 R R

4 = m f1.15 R R

41.15

= m f3.478 R R


14 = fR 1.75 3.478

fR = 14 6.087 7.913%

BR = 7.913 0.82 3.478 = 10.76%


B m cBβ σ +σ

26 = 2 2B mβ σ +5


9SANJAY SARAF SIR

36 = 2 2m0.82 σ +5

36 - 5 = 2m0.6724σ

310.6724

= 2mσ

46.10 = 2mσ

2A = 246.10 1.75 12 153.182C = 246.10 0.60 18 34.6

A = 12.38%

C = 5.88%

PROBLEM - 7

Given below are the risk estimates for two stocks X and Y:

Stock Covariance withthe market

ExpectedReturn

Firm SpecificVariance

X 435.6(%)2 14% 625(%)2

Y 580.8(%)2 18% 1,225(%)2

The risk-free rate of return is 6%, and the standard deviation of the returns on themarket index is 22%.

An investor is considering the following two alternatives for investing in the abovestocks:

i. Place equal proportion of his money in both the stocks.

ii. Place 25% of his money in Stock X, 40% of his money in stock Y and place hisremaining money in the risk-free T-Bills.

You are required to

a. Compute the total risk associated with stocks X and Y.

b. Calculate the expected return and the total risk associated with both thealternatives. Which alternative should the investor prefer if he is consideringcoefficient of variation as the benchmark?


9SANJAY SARAF SIR

36 = 2 2m0.82 σ +5

36 - 5 = 2m0.6724σ

310.6724

= 2mσ

46.10 = 2mσ

2A = 246.10 1.75 12 153.182C = 246.10 0.60 18 34.6

A = 12.38%

C = 5.88%

PROBLEM - 7



ExpectedReturn


X 435.6(%)2 14% 625(%)2

Y 580.8(%)2 18% 1,225(%)2





You are required to




9SANJAY SARAF SIR

36 = 2 2m0.82 σ +5

36 - 5 = 2m0.6724σ

310.6724

= 2mσ

46.10 = 2mσ

2A = 246.10 1.75 12 153.182C = 246.10 0.60 18 34.6

A = 12.38%

C = 5.88%

PROBLEM - 7



ExpectedReturn


X 435.6(%)2 14% 625(%)2

Y 580.8(%)2 18% 1,225(%)2





You are required to




10SANJAY SARAF SIR

Solution :a.

Stocks 2

Cov Stock Mktβ

σ m

SR = 2 2m UR TR

X 0.9 392.04 625 1017.04Y 1.2 696.96 1225 1921.95

b. Alternative 1 : Place equal proportion of his money in both the stocks.2 2 2 2 2 2 P x x y x y 2X Y m

= 254.26 +480.49 +261.362 299611P . %

31 56P . % , 14 18162

PE R %

31 56 100 197 2516

SD .CV . %mean

Alternative 2 : Place 25% of his money in Stock X, 40% of his money in stock Yand place his remaining money in the risk-free T-Bills.

2 22 20 25 1017 04 0 4 192116 2 0 25 0 4 0 9 1 2 22 P . . . . . . . .

63 565 307 51 104 54 . . .= 475.61521 81P . %

0 25 14 0 4 18 0 35 6 PE R . . .

= 12.8%21 81 170 3912 8

SD .CV . %mean .

Alternative 2 should be choosen.


10SANJAY SARAF SIR

Solution :a.

Stocks 2

Cov Stock Mktβ

σ m

SR = 2 2m UR TR

X 0.9 392.04 625 1017.04Y 1.2 696.96 1225 1921.95


= 254.26 +480.49 +261.362 299611P . %

31 56P . % , 14 18162

PE R %

31 56 100 197 2516

SD .CV . %mean


2 22 20 25 1017 04 0 4 192116 2 0 25 0 4 0 9 1 2 22 P . . . . . . . .

63 565 307 51 104 54 . . .= 475.61521 81P . %

0 25 14 0 4 18 0 35 6 PE R . . .

= 12.8%21 81 170 3912 8

SD .CV . %mean .



10SANJAY SARAF SIR

Solution :a.

Stocks 2

Cov Stock Mktβ

σ m

SR = 2 2m UR TR

X 0.9 392.04 625 1017.04Y 1.2 696.96 1225 1921.95


= 254.26 +480.49 +261.362 299611P . %

31 56P . % , 14 18162

PE R %

31 56 100 197 2516

SD .CV . %mean


2 22 20 25 1017 04 0 4 192116 2 0 25 0 4 0 9 1 2 22 P . . . . . . . .

63 565 307 51 104 54 . . .= 475.61521 81P . %

0 25 14 0 4 18 0 35 6 PE R . . .

= 12.8%21 81 170 3912 8

SD .CV . %mean .



11SANJAY SARAF SIR

PROBLEM - 8

Consider the following data for two companies and the market:

Company/Market Beta StandardDeviation (%)

Covariance withSensex (%)2

Zee Teleflims N.A. 45 205Padmalay Teleflims 1.2 40 N.A

Teleflims Sensex 1.0 15 225

Further it is gathered that risk - free interest is 7%. Considering the assumptions ofregression (Characteristic) line hold good you are required to find.

a.i. Beta of Zee Teleflims.ii. Covariance of return on Padmalay Teleflims with that of return on sensex.

b. The coefficients of correlation between:i. Return on Zee Teleflims and return on sensex.ii. Return on Padmalaya Teleflims and return on Sensex.

c. The variance of the portfolio formed using Zee Telefilms and Padmalya Teleflimsin the proportion of 2/3 and 1/3 respectively.

d. Whether the unsystematic risk of the portfolio is less than individual companies?(β of portfolio is weighted average betas of underlying stocks).

Solution :

a.

i.

zzz

Cov Zee,MVar M

β

0.0205 0.910.0225


11SANJAY SARAF SIR

PROBLEM - 8











Solution :

a.

i.

zzz

Cov Zee,MVar M

β

0.0205 0.910.0225


11SANJAY SARAF SIR

PROBLEM - 8











Solution :

a.

i.

zzz

Cov Zee,MVar M

β

0.0205 0.910.0225


12SANJAY SARAF SIR

ii. PadCov Pad, M Var Mβ

= 1.2 × 0.0225= 0.027 i.e. 270 (%2)

b.

i.

zzzzee M

Cov zee, MMρ

σ σ

0.0205

0.45 0.0225= 0.304

ii.

pad.Pad M

Cov Pad.MMρ

σ σ

0.027

0.40 0.0225= 0.45

c. Var (Portfolio) 2 2 2zee zee Pad zee padW W 2W , W Cov Zee, Padσ

see zee2W ; 0.453σ

see pad1W ; 0.403σ

From the assumption of characteristic (Regression) line we get

zee padCov zee, pad Var Mβ β

= 0.91 × 1.2 × 0.0225= 0.025 i.e, 250 (%2)

Variance (Portfolio)

2 22 22 10.45 0.40

3 3

2 12 0.0253 3

= 0.119 i.e, 1190 (%2)


12SANJAY SARAF SIR


= 1.2 × 0.0225= 0.027 i.e. 270 (%2)

b.

i.

zzzzee M

Cov zee, MMρ

σ σ

0.0205

0.45 0.0225= 0.304

ii.

pad.Pad M

Cov Pad.MMρ

σ σ

0.027

0.40 0.0225= 0.45


see zee2W ; 0.453σ

see pad1W ; 0.403σ



= 0.91 × 1.2 × 0.0225= 0.025 i.e, 250 (%2)


2 22 22 10.45 0.40

3 3

2 12 0.0253 3

= 0.119 i.e, 1190 (%2)


12SANJAY SARAF SIR


= 1.2 × 0.0225= 0.027 i.e. 270 (%2)

b.

i.

zzzzee M

Cov zee, MMρ

σ σ

0.0205

0.45 0.0225= 0.304

ii.

pad.Pad M

Cov Pad.MMρ

σ σ

0.027

0.40 0.0225= 0.45


see zee2W ; 0.453σ

see pad1W ; 0.403σ



= 0.91 × 1.2 × 0.0225= 0.025 i.e, 250 (%2)


2 22 22 10.45 0.40

3 3

2 12 0.0253 3

= 0.119 i.e, 1190 (%2)


13SANJAY SARAF SIR

d. Unsystematic Risk of Zee Telefilms

2 2zee1 Mρ σ

2 21 0.304 0.45

= 0.184 i.e 1840 (%2)

Unsystematic Risk of Padmalay Telefilms

2 2pad.M Pad1 ρ σ

2 21 0.45 0.40

= 0.128 i.e, 1280 (%2)

portfolio zee pad2 13 3

β β β

2 10.91 1.2 1.0073 3

port M

Cov Port.M.Port.Mρ

σ σ

portM

Portσβσ

0.02251.007 0.4380.119

Unsystematic risk of portfolio

2 2port port1 .Mρ σ

21 0.438 0.119

= 0.096 i.e 960 (%2)

Therefore, we find that the unsystematic Risk of the portfolio is less than that ofindividual stocks. From the result it can be implied that because of constitution ofportfolio unsystematic return reduces.


13SANJAY SARAF SIR


2 2zee1 Mρ σ

2 21 0.304 0.45

= 0.184 i.e 1840 (%2)


2 2pad.M Pad1 ρ σ

2 21 0.45 0.40

= 0.128 i.e, 1280 (%2)


β β β

2 10.91 1.2 1.0073 3

port M

Cov Port.M.Port.Mρ

σ σ

portM

Portσβσ

0.02251.007 0.4380.119



21 0.438 0.119

= 0.096 i.e 960 (%2)



13SANJAY SARAF SIR


2 2zee1 Mρ σ

2 21 0.304 0.45

= 0.184 i.e 1840 (%2)


2 2pad.M Pad1 ρ σ

2 21 0.45 0.40

= 0.128 i.e, 1280 (%2)


β β β

2 10.91 1.2 1.0073 3

port M

Cov Port.M.Port.Mρ

σ σ

portM

Portσβσ

0.02251.007 0.4380.119



21 0.438 0.119

= 0.096 i.e 960 (%2)



14SANJAY SARAF SIR

PROBLEM - 9

Mr. A. Rathi is testing the weak form efficient market hypothesis on the Indian stockmarket. For this he has collected the data on a leading market index for the last 15trading days. This is given below:

Trading day Market Index123456789

101112131415

450045504400435043004330440044454440437043804365450045604600

You are required to perform a runs test and determine the independence of data at10% level of significance.


14SANJAY SARAF SIR

PROBLEM - 9



101112131415

450045504400435043004330440044454440437043804365450045604600



14SANJAY SARAF SIR

PROBLEM - 9



101112131415

450045504400435043004330440044454440437043804365450045604600



15SANJAY SARAF SIR

Solution :Trading day Market Index

1 45002 4550 + 1 8n

3 4400 - 2 6n

4 4350 -5 4300 - r 76 4330 +7 4400 +8 4445 +9 4440 -

10 4370 -11 4380 +12 4365 -13 4500 +14 4560 +15 4600 +

1 2

1 2

2 1 7 86 n n .n n

1 2

1 21 76

1

.n n

Case 1 10 % significance (t = 1.771)4 74 C t . 10 98 U t .

r i.e 7 lies between 4.74 & 10.98Market is efficient in the weak form.


15SANJAY SARAF SIR


1 45002 4550 + 1 8n

3 4400 - 2 6n

4 4350 -5 4300 - r 76 4330 +7 4400 +8 4445 +9 4440 -

10 4370 -11 4380 +12 4365 -13 4500 +14 4560 +15 4600 +

1 2

1 2

2 1 7 86 n n .n n

1 2

1 21 76

1

.n n




15SANJAY SARAF SIR


1 45002 4550 + 1 8n

3 4400 - 2 6n

4 4350 -5 4300 - r 76 4330 +7 4400 +8 4445 +9 4440 -

10 4370 -11 4380 +12 4365 -13 4500 +14 4560 +15 4600 +

1 2

1 2

2 1 7 86 n n .n n

1 2

1 21 76

1

.n n



Documents

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