Risk and Return
Normal distribution and standardization
Obs (CP) Mean Obs - Mean (Obs – Mean)/SD
355
515.5
-160.5 -0.65748
385
515.5
-130.5 -0.53458
410
515.5
-105.5 -0.43217
510
515.5
-5.5 -0.02253
84
515.5
-431.5 -1.7676
304
515.5
-211.5 -0.86639
853
515.5
337.5 1.382541
770
515.5
254.5 1.042539
864
515.5
348.5 1.427602
620
515.5
104.5 0.428076
Mean value = = 515.5 Mean value=0
SD = = 244.11 SD= 1
If X is a random variable and has a normal distribution with mean µ and standard deviation σ,
then the Empirical Rule says the following:
About 68% of the x values lie between –1σ and +1σ of the mean µ (within one standard
deviation of the mean).
About 95% of the x values lie between –2σ and +2σ of the mean µ (within two standard
deviations of the mean).
About 99.7% of the x values lie between –3σ and +3σ of the mean µ(within three standard
deviations of the mean).
The z-scores for +1σ and –1σ are +1 and –1, respectively.
The z-scores for +2σ and –2σ are +2 and –2, respectively.
The z-scores for +3σ and –3σ are +3 and –3 respectively.
If a stock has a mean return of 8% and a Std. Dev of 10%. Find its max and min
value upto 3 std dev assuming that the data points are distributed normally
Mean = μ = 0.08
Std dev = σ = 0.1
Formula for min and max value μ ± zσ
Min and max values for 1 std dev
μ ± zσ = μ ± 1σ = 0.08 ± 1*0.1 = -0.02 to 0.18 (About 68% of the x values lie Between -0.02 and 0.18)
Min and max values for 2 std dev
μ ± zσ = μ ± 2σ = 0.08 ± 2*0.1 = -0.12 to 0.28 (About 95% of the x values lie Between -0.02 and 0.18)
Min and max values for 3 std dev
μ ± zσ = μ ± 3σ = 0.08 ± 3*0.1 = -0.22 to 0.38 (About 99.7% of the x values lie Between -0.02 and 0.18)
What is the probability that a stock with mean return of 8% and standard
deviation of 10% would yield a return of 6% or less.
The probability that return is less than 6% [P(x<6)=0.4207]
What is the probability that a stock with mean return of 8% and standard
deviation of 10% would yield a return of 15% or less.
The probability that return is less than 15% [P(x<15)=0.7580]
What is the probability that a stock with mean return of 8% and standard deviation of 10% would
yield a return between 12% and 18%
[P(x<12)=0.6554]
[P(x<18)=0.8413]
[P(12<x<18) = 0.8413 – 0.6554 = 0.1859]
Covariance and correlation
• Covariance: Co-movement between two
assets
• Correlation: Standardized version of
covariance
Covariance (Historical data)
Given below are the closing stock prices of two companies. Find
out the covariance between them.
Year Raincoat comp Sunglass comp
2017 100 300
2018 90 330
2019 80 350
2020 75 360
Covariance (Probabilistic data)
You bought a share of Raincoat company for Rs 120 and one
share of Sunglass company for Rs 280. Under the current
scenario, three probabilities of share price movements for the
companies are likely
Prob RC SC
0.3 100 300
0.2 90 330
0.5 80 350
Beta
Beta (Sensitivity) Systematic risk
Historical data
Beta (Sensitivity) Systematic risk
Probabilistic data

risk and return fm

  • 1.
  • 2.
    Normal distribution andstandardization
  • 3.
    Obs (CP) MeanObs - Mean (Obs – Mean)/SD 355 515.5 -160.5 -0.65748 385 515.5 -130.5 -0.53458 410 515.5 -105.5 -0.43217 510 515.5 -5.5 -0.02253 84 515.5 -431.5 -1.7676 304 515.5 -211.5 -0.86639 853 515.5 337.5 1.382541 770 515.5 254.5 1.042539 864 515.5 348.5 1.427602 620 515.5 104.5 0.428076 Mean value = = 515.5 Mean value=0 SD = = 244.11 SD= 1
  • 4.
    If X isa random variable and has a normal distribution with mean µ and standard deviation σ, then the Empirical Rule says the following: About 68% of the x values lie between –1σ and +1σ of the mean µ (within one standard deviation of the mean). About 95% of the x values lie between –2σ and +2σ of the mean µ (within two standard deviations of the mean). About 99.7% of the x values lie between –3σ and +3σ of the mean µ(within three standard deviations of the mean). The z-scores for +1σ and –1σ are +1 and –1, respectively. The z-scores for +2σ and –2σ are +2 and –2, respectively. The z-scores for +3σ and –3σ are +3 and –3 respectively.
  • 5.
    If a stockhas a mean return of 8% and a Std. Dev of 10%. Find its max and min value upto 3 std dev assuming that the data points are distributed normally Mean = μ = 0.08 Std dev = σ = 0.1 Formula for min and max value μ ± zσ Min and max values for 1 std dev μ ± zσ = μ ± 1σ = 0.08 ± 1*0.1 = -0.02 to 0.18 (About 68% of the x values lie Between -0.02 and 0.18) Min and max values for 2 std dev μ ± zσ = μ ± 2σ = 0.08 ± 2*0.1 = -0.12 to 0.28 (About 95% of the x values lie Between -0.02 and 0.18) Min and max values for 3 std dev μ ± zσ = μ ± 3σ = 0.08 ± 3*0.1 = -0.22 to 0.38 (About 99.7% of the x values lie Between -0.02 and 0.18)
  • 6.
    What is theprobability that a stock with mean return of 8% and standard deviation of 10% would yield a return of 6% or less. The probability that return is less than 6% [P(x<6)=0.4207]
  • 7.
    What is theprobability that a stock with mean return of 8% and standard deviation of 10% would yield a return of 15% or less. The probability that return is less than 15% [P(x<15)=0.7580]
  • 8.
    What is theprobability that a stock with mean return of 8% and standard deviation of 10% would yield a return between 12% and 18% [P(x<12)=0.6554] [P(x<18)=0.8413] [P(12<x<18) = 0.8413 – 0.6554 = 0.1859]
  • 9.
    Covariance and correlation •Covariance: Co-movement between two assets • Correlation: Standardized version of covariance
  • 10.
    Covariance (Historical data) Givenbelow are the closing stock prices of two companies. Find out the covariance between them. Year Raincoat comp Sunglass comp 2017 100 300 2018 90 330 2019 80 350 2020 75 360
  • 11.
    Covariance (Probabilistic data) Youbought a share of Raincoat company for Rs 120 and one share of Sunglass company for Rs 280. Under the current scenario, three probabilities of share price movements for the companies are likely Prob RC SC 0.3 100 300 0.2 90 330 0.5 80 350
  • 12.
  • 13.
    Beta (Sensitivity) Systematicrisk Historical data
  • 14.
    Beta (Sensitivity) Systematicrisk Probabilistic data