Expected return and standard deviation:

a. Johnson & Johnson (JNJ) is trading at 123.64 (5/12/2017 close). JNJ is a large health care conglomerate. It has done well over the last couple of years and you think it will continue to do well. After careful analysis you conclude that in one year the price will be (90, 105, 125, 155, 175) with associated probabilities of (0.1, 0.2, 0.4, 0.2, 0.1). Looking at the company’s past record you project that JNJ will pay a dividend of 3.40 (four quarterly dividends of 0.85).
(i) What is the expected return of JNJ stock?
(ii) Calculate the standard deviation of the return of JNJ stock (remember that you are using probabilities to do this, not historical data).

b. In the second sheet of the Excel file PS4 you will find historical data for IBM and CVX returns. (i) Calculate the sample expected return and sample standard deviation for both. (ii) What is the standard error of the sample average in each case? What is the +/- 2 standard error confidence interval of the sample average?

Expected Return:

Expected Price in one year

= (90x 0.10)+ (105x 0.20)+ (125x 0.40)+ (155x 0.20)+ (175x 0.10)

= 128.5 Expected dividends in one year

= 3.40

Expected Return = 123.64 Expected Return=[ (Expected Price + Expected Dividends)] / Current Price= >[ (128.50 + 3.40)–123.64]/123.64= 0.066806859 or 6.68%

Standard Deviation: To measure the volatility, the estimated return must be determined with each price point.

Expected return = = [(Expected Price + Expected dividends) – Current Price] / Current Price

Current Price     Expected Price     Dividend       Expected Return

123.64                   90                          3.4                     -24.46%

123.64                   105                         3.4                     -12.33%

123.64                   125                         3.4                      3.85%

123.64                   155                         3.4                      28.11%

123.64                   175                         3.4                      44.29%

Variance = [(-0.2446 – 0.0668)^2 x 0.10] + [(-0.1233 – 0.0668)^2 x 0.20] + [(0.0385 – 0.0668)^2 x 0.40] + [(0.2811 – 0.0668)^2 x 0.20] + [(0.4429 – 0.0668)^2 x 0.10] = 0.040574089

Standard Deviation = (0.040574089)1/2 = 0.20143011 or 20.14%

Explanation:

answer; the correct answer should be (demonstrations); good luck