Question

Tyler is the owner of a small company. He has hired Marla, a business consultant, to help him expand his company. Marla will conduct some basic analyses on Tyler's company before giving any recommendations In one analysis, she considered the log-daily sales revenue (that is, she has taken the past daily sales revenue amounts and taken the natural log of them). She believes that the log-daily sales follows a Normal distribution with parameters u = 1 and o2 0.0025. Assuming Marla is correct, answer the following State the distribution for the daily sales revenue a) [ mark] b) Calculate the mean and standard deviation for the daily sales revenue [2 marks] c) What is the probability that Tyler's sales revenue tomorrow will exceed $2.95? [3 marks] d) What is the 10% VaR for Tyler's daily sales revenue? [3 marks] e) If Marla used a t distribution instead of a Normal distribution for the log-daily sales, how do you think it would affect your answer to part d)? Note: you do NOT need to perform any further calculations [ mark

Answer 1

(a)

Sales revenue on t^th day = R_t

ln R_t$\sim$ N( $\mu$ , $\sigma$ ² )   $\Rightarrow$ R_t$\sim$ logN($\mu$, $\sigma$ ² )  

i.e. R_t has log-normal distribution with parameters $\mu$ = 1 and $\sigma$ ² = 0.0025

(b)

E[R_t] = e$\mu$^{+ 0.5$\sigma$^2} = e^{1 + 0.5x0.0025} = $2.7216818

Var[R_t] = e²$\mu$^{+$\sigma$^2} (e^{$\sigma$^2} - 1) = e^{2x1 + 0.0025}(e^0.0025 - 1) =$² 0.018542047

(c)

P[R_t > 2.95] = P[ln R_t > ln 2.95] = P[ N( $\mu$ , $\sigma$ ² ) > ln 2.95] = 1 -  P[ N( $\mu$ , $\sigma$ ² ) <  ln 2.95] = 1 - $\Phi$ {(ln 2.95 -  $\mu$) / $\sigma$ }

= 1 - $\Phi$ {(ln 2.95 - 1) / 0.0025^0.5}-1 = 1 - $\Phi$ (1.6361) = 1-0.94909 = 0.05091

(d)

P[R_t < y] = 1 - 0.10

$\Rightarrow$ P[ ln R_t < ln y ] = 0.90

$\Rightarrow$ P[ N( $\mu$ , $\sigma$ ² ) < ln y ] = 0.90

$\Rightarrow$ P[ N(0, 1) < ( ln y - $\mu$ ) / $\sigma$ ] = 0.90

$\Rightarrow$$\Phi$( ( ln y - $\mu$ ) / $\sigma$ ) = 0.90

$\Rightarrow$ ( ln y - $\mu$ ) / $\sigma$ = $\Phi$ ^-1(0.90)

$\Rightarrow$ (ln y - 1) / 0.0025^0.5 = 1.2816

$\Rightarrow$ y = 1.06408

$\Rightarrow$ VaR(R_t) = - y = -1.06408