CM2A 2022 04 Q5 ii) — am I going nuts?

Hi there!

I compared my answer to the Examiner’s Report. The two markers gave me respectively 7.5/12 and 4.5/12. But I cannot find any error in my answer to warrant such harsh marking—the only error I made was not squaring my standard deviation to arrive at a variance.
I know it is somewhat tedious, but this has been driving me crazy, I cannot see how I lost so many marks.

Please can someone just help me look at it and let me know.
It is also vital for when I rewrite that I know what was so important that I left out.
Thank you so much!

I have copied my solution below.
Here is a link to a photo of their solution, so that it isn’t tricky to find the ER.
https://share.icloud.com/photos/01b3rq_GjjU9GGE_ceUlahSVw

Q5 ii)
a)

by independence

EV = 500*E[(Iti 1)*(I+i 2)*(1+i 3)*.. *(Iti 15)]

=500*1.035^15=837.6744153761

variance:

(Iti 1)*(I+i 2)*(1+i 3)*...*(Iti 15)

= prod(t=1, t=15):[1+i_t]

=log(prod(t=1, t=15):[1+i_t]) = sum(t=1,t=15): [log(Iti_t)]

since {¡ t} are independent:

sum(t=1,t=15):[log(Iti_t)] ~ N(15*м, 15*0^2)

prod(t=1, +=15):[1+i_t] ~ logN(15*M, 15*6^2)

var[prod(t=1, t=15):[1+i_t]] = exp(2*15*0.0342+15*0.0193)*

(exp(15*0.0193)-1)= 1.2512457763

500^2*1.2512457763=312 811.444075

sd = sqrt(312 811.444075)= 559.295489053

b)

500*1.04^10 = 740.1221424592

EV=

740.1221424592*1.005^5*0.15+

740.1221424592*1.01^5*0.25+

740.1221424592*1.045^5*0.4+

740.1221424592*1.07^5*0.2

=884.8333

let X = value of the investment at time=15 years

variance = E[X^21 - E[X1^2

E[X^21

=(740.1221424592*1.005^5^2*0.15+

(740.1221424592*1.01^5)^2*0.25+

(740.1221424592*1.045^5)^2*0.4+

(740.1221424592*1.07^5)^2*0.2

=793430.1

variance = 793430.1 - 884.8333^2 = £^2 10 500.13121111 =

(£102.4701479023)^2

sd=sqrt(variance) ~= £102.47

 

Ok thank you anyway .