Question

Below are returns on the stock A and S&P500 index. All numbers are in decimals (-0.0222 is equivalent to -2.22%).

A	S_P500
-0.0222	0.0032
-0.0048	-0.0058
0.1333	0.0434
0.0765	0.1081
-0.0161	-0.0121
0.1250	0.1400
0.0145	0.0368
-0.0475	-0.0454
0.0430	0.0577
-0.0260	-0.1374
0.0071	0.0064
0.0249	0.0186
0.0850	0.0215
-0.0624	-0.0752
0.0933	0.0365
0.0456	0.0528
-0.0632	-0.0131
0.0450	0.0009
0.0200	0.0017
0.0280	0.0985

Assuming normal distribution, what is the probability that the next S&P500 return will be greater than 0?

Please write an answer in decimals. For example, 12% would be 0.12.
Also, round your answer to the second decimal.

Answer 1

We are assuming that the distribution of the return of the S&P 500 is normal distribution, so we will convert it into a standard normal distribution.

S&P 500	(X-μs&p 500)2
0.0032	0.000186459
-0.0058	0.000513249
0.0434	0.000704637
0.1081	0.00832565
-0.0121	0.000838392
0.14	0.015164691
0.0368	0.000397803
-0.0454	0.003875685
0.0577	0.001668314
-0.1374	0.023794605
0.0064	0.000109307
0.0186	3.04502E-06
0.0215	2.1576E-05
-0.0752	0.008474123
0.0365	0.000385926
0.0528	0.001292043
-0.0131	0.000897302
0.0009	0.000254562
0.0017	0.000229674
0.0985	0.006665906
μs&p 500= 0.016855	sum = 0.073803

S&P 500	(X-μs&p 500)2
0.0032	0.000186459
-0.0058	0.000513249
0.0434	0.000704637
0.1081	0.00832565
-0.0121	0.000838392
0.14	0.015164691
0.0368	0.000397803
-0.0454	0.003875685
0.0577	0.001668314
-0.1374	0.023794605
0.0064	0.000109307
0.0186	3.04502E-06
0.0215	2.1576E-05
-0.0752	0.008474123
0.0365	0.000385926
0.0528	0.001292043
-0.0131	0.000897302
0.0009	0.000254562
0.0017	0.000229674
0.0985	0.006665906
μs&p 500= 0.016855	sum = 0.073803

σ_{s&p 500} = [0.073803/(20-1)]^1/2 = 0.062324680

we can calculate the standard deviation from the given sample of S&P 500 return using the formula =STDEV.S(Range of S&P 500)

σ_{s&p 500} = 0.0623247

Let X be the next return of S&P 500.

we need to calculate the probability that X > 0. i.e.,

where μ is the mean of the sample and σ is the standard deviation of the sample of S&P 500

Z = (X-μ)/σ

hence, we need to calculate that P(Z>(0-μ_{s&p 500})/σ_{s&p 500}) = P[Z> (0 - 0.016855)/0.06232468] = P(Z> -0.2704) = 1 - P(Z < -0.2704)

From Z-distribution table we can see that P(Z < -0.2704) = 0.39342627. We can also calculate this using the excel function =NORM.S.DIST(-0.2704,TRUE) =0.39342627

Hence P(Z> -0.2704) = 1 - P(Z < -0.2701) = 1-0.39342627 = 0.60657373

Answer -> 0.61