Answer to Question #285907 in Finance for kelly

Question #285907

Given the end-of-month prices for stock X for January 2020 through June 2020,

Month End-of-Month Price

January 2020 15.25

February 2020 16.10

March 2020 16.20

April 2020 15.80

May 2020 15.85

June 2020 16.30

calculate

a.        Monthly holding-period return

b.              Average Return

c.               Standard Deviation


1
Expert's answer
2022-01-14T10:15:10-0500

(b) We shall calculate the average return bybthe formula

Average return =averageinvestmentinitialinvestment=\frac{average \hspace{0.1cm}investment}{initial \hspace{0.1cm}investment}



The average investment from January through to June is given by

=95.56=\frac{95.5}{6}

=15.92=15.92


Now,

Averagereturn=15.9215.25Average \hspace{0.1cm}return=\frac{15.92}{15.25}

=1.04=1.04



(C) To calculate the standard deviation

S.D=(xxˉ)²nS.D=\sqrt{\frac{\sum(x-\bar{x})²}{n}}


Where xˉ\bar{x} Is the average value (the mean) in question (b) which is

xˉ=15.92\bar{x}=15.92


(xxˉ)²=(15.2515.92)²+(16.1015.92)²+(16.2015.92)²+(15.8015.92)²+(15.8515.92)²+(16.3015.92)²\sum({x-\bar{x}})²=(15.25-15.92)² +(16.10-15.92)²+(16.20-15.92)²+(15.80-15.92)²+(15.85-15.92)²+(16.30-15.92)²


(xxˉ)=0.4489+0.0324+0.0729+0.0144+0.0049+0.1444\sum({x-\bar{x}})=0.4489+0.0324+0.0729+0.0144+0.0049+0.1444


(xxˉ)²=0.7179\sum({x-\bar{x}})²=0.7179


Therefore,

S.D=(xxˉ)²nS.D=\sqrt{\frac{\sum(x-\bar{x})²}{n}}


S.D=0.71796S.D=\sqrt{\frac{0.7179}{6}}


S.D=0.11965S.D=0.3459S.D=\sqrt{0.11965}\\S.D=0.3459


HENCE,

(a)the monthly holding -period return is

0.06%

(b)the average return is

1.04%

(c)the standard deviation is

0.3459


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