Probability distribution of the future stock price
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I have this exercise to do:
Consider a stock with an initial price of $40, an expected return of 16% per annum and a volatility of 20% per annum. Find the probability distribution of the future stock price in 6 months' time. Determine the 95% confidence of interval.
$S_0=40$
$mu=0.16$
$sigma=0.20$
$T=0.5$
Using the lognormal property of stock prices $(ln{S_T}approxvarphileft[ln{S_0}+left(mu-frac{sigma^2}{2}right)T, sigma^2Tright]$, I have $ln{S_T}approxvarphileft[3.759, 0.02right]$.
The results of the textbook says "there is a 95% probability that a normally distributed variable has a value within 1.96 standard deviaton of its mean."
Viewing another topic from this forum, I found this possible solution: $P[lnS_T>ln95]=1−P[lnS_T<ln95]=1−P[Z<z=frac{ln95-3.759}{sqrt{0.02}}]$
If I want to compute $z$, as result I will have $z=5.62$. Thus, $N(5.62)=1$ (is not possible for this exercise).
Supposing a 95% c.i. I can't solve the problem. Is anyone can explain me how can I approach it?
I would be beyond thankful for any help and advice.
probability normal-distribution finance
asked Dec 7 '18 at 19:17
Martina MartyMartina Marty
225
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add a comment |
$begingroup$
I have this exercise to do:
Consider a stock with an initial price of $40, an expected return of 16% per annum and a volatility of 20% per annum. Find the probability distribution of the future stock price in 6 months' time. Determine the 95% confidence of interval.
$S_0=40$
$mu=0.16$
$sigma=0.20$
$T=0.5$
Using the lognormal property of stock prices $(ln{S_T}approxvarphileft[ln{S_0}+left(mu-frac{sigma^2}{2}right)T, sigma^2Tright]$, I have $ln{S_T}approxvarphileft[3.759, 0.02right]$.
The results of the textbook says "there is a 95% probability that a normally distributed variable has a value within 1.96 standard deviaton of its mean."
Viewing another topic from this forum, I found this possible solution: $P[lnS_T>ln95]=1−P[lnS_T<ln95]=1−P[Z<z=frac{ln95-3.759}{sqrt{0.02}}]$
If I want to compute $z$, as result I will have $z=5.62$. Thus, $N(5.62)=1$ (is not possible for this exercise).
Supposing a 95% c.i. I can't solve the problem. Is anyone can explain me how can I approach it?
I would be beyond thankful for any help and advice.
probability normal-distribution finance
asked Dec 7 '18 at 19:17
Martina MartyMartina Marty
225
$endgroup$
$begingroup$
"$textrm{The results of the textbook says "there is a 95% probability that a normally distributed ...}$" This statement just means that $P(-1.96 leq Z leq 1.96)=0.95$, where $Zsim mathcal N(0,1)$
$endgroup$
– callculus
Dec 7 '18 at 19:32
1
$begingroup$
Thank you, now I've understood my mistake
$endgroup$
– Martina Marty
Dec 7 '18 at 19:41
$begingroup$
That makes me happy. You´re welcome.
$endgroup$
– callculus
Dec 7 '18 at 19:43
add a comment |
$begingroup$
I have this exercise to do:
Consider a stock with an initial price of $40, an expected return of 16% per annum and a volatility of 20% per annum. Find the probability distribution of the future stock price in 6 months' time. Determine the 95% confidence of interval.
$S_0=40$
$mu=0.16$
$sigma=0.20$
$T=0.5$
Using the lognormal property of stock prices $(ln{S_T}approxvarphileft[ln{S_0}+left(mu-frac{sigma^2}{2}right)T, sigma^2Tright]$, I have $ln{S_T}approxvarphileft[3.759, 0.02right]$.
The results of the textbook says "there is a 95% probability that a normally distributed variable has a value within 1.96 standard deviaton of its mean."
Viewing another topic from this forum, I found this possible solution: $P[lnS_T>ln95]=1−P[lnS_T<ln95]=1−P[Z<z=frac{ln95-3.759}{sqrt{0.02}}]$
If I want to compute $z$, as result I will have $z=5.62$. Thus, $N(5.62)=1$ (is not possible for this exercise).
Supposing a 95% c.i. I can't solve the problem. Is anyone can explain me how can I approach it?
I would be beyond thankful for any help and advice.
probability normal-distribution finance
asked Dec 7 '18 at 19:17
Martina MartyMartina Marty
225
$endgroup$
I have this exercise to do:
Consider a stock with an initial price of $40, an expected return of 16% per annum and a volatility of 20% per annum. Find the probability distribution of the future stock price in 6 months' time. Determine the 95% confidence of interval.
$S_0=40$
$mu=0.16$
$sigma=0.20$
$T=0.5$
Using the lognormal property of stock prices $(ln{S_T}approxvarphileft[ln{S_0}+left(mu-frac{sigma^2}{2}right)T, sigma^2Tright]$, I have $ln{S_T}approxvarphileft[3.759, 0.02right]$.
The results of the textbook says "there is a 95% probability that a normally distributed variable has a value within 1.96 standard deviaton of its mean."
Viewing another topic from this forum, I found this possible solution: $P[lnS_T>ln95]=1−P[lnS_T<ln95]=1−P[Z<z=frac{ln95-3.759}{sqrt{0.02}}]$
If I want to compute $z$, as result I will have $z=5.62$. Thus, $N(5.62)=1$ (is not possible for this exercise).
Supposing a 95% c.i. I can't solve the problem. Is anyone can explain me how can I approach it?
I would be beyond thankful for any help and advice.
probability normal-distribution finance
probability normal-distribution finance
asked Dec 7 '18 at 19:17
Martina MartyMartina Marty
225
asked Dec 7 '18 at 19:17
Martina MartyMartina Marty
225
edited Dec 7 '18 at 19:30
Martina Marty
asked Dec 7 '18 at 19:17
Martina MartyMartina Marty
225
asked Dec 7 '18 at 19:17
Martina MartyMartina Marty
225
asked Dec 7 '18 at 19:17
Martina MartyMartina Marty
225
225
$begingroup$
"$textrm{The results of the textbook says "there is a 95% probability that a normally distributed ...}$" This statement just means that $P(-1.96 leq Z leq 1.96)=0.95$, where $Zsim mathcal N(0,1)$
$endgroup$
– callculus
Dec 7 '18 at 19:32
1
$begingroup$
Thank you, now I've understood my mistake
$endgroup$
– Martina Marty
Dec 7 '18 at 19:41
$begingroup$
That makes me happy. You´re welcome.
$endgroup$
– callculus
Dec 7 '18 at 19:43
add a comment |
$begingroup$
"$textrm{The results of the textbook says "there is a 95% probability that a normally distributed ...}$" This statement just means that $P(-1.96 leq Z leq 1.96)=0.95$, where $Zsim mathcal N(0,1)$
$endgroup$
– callculus
Dec 7 '18 at 19:32
1
$begingroup$
Thank you, now I've understood my mistake
$endgroup$
– Martina Marty
Dec 7 '18 at 19:41
$begingroup$
That makes me happy. You´re welcome.
$endgroup$
– callculus
Dec 7 '18 at 19:43
$begingroup$
"$textrm{The results of the textbook says "there is a 95% probability that a normally distributed ...}$" This statement just means that $P(-1.96 leq Z leq 1.96)=0.95$, where $Zsim mathcal N(0,1)$
$endgroup$
– callculus
Dec 7 '18 at 19:32
$begingroup$
"$textrm{The results of the textbook says "there is a 95% probability that a normally distributed ...}$" This statement just means that $P(-1.96 leq Z leq 1.96)=0.95$, where $Zsim mathcal N(0,1)$
$endgroup$
– callculus
Dec 7 '18 at 19:32
1
1
$begingroup$
Thank you, now I've understood my mistake
$endgroup$
– Martina Marty
Dec 7 '18 at 19:41
$begingroup$
Thank you, now I've understood my mistake
$endgroup$
– Martina Marty
Dec 7 '18 at 19:41
$begingroup$
That makes me happy. You´re welcome.
$endgroup$
– callculus
Dec 7 '18 at 19:43
$begingroup$
That makes me happy. You´re welcome.
$endgroup$
– callculus
Dec 7 '18 at 19:43
add a comment |
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$begingroup$
"$textrm{The results of the textbook says "there is a 95% probability that a normally distributed ...}$" This statement just means that $P(-1.96 leq Z leq 1.96)=0.95$, where $Zsim mathcal N(0,1)$
$endgroup$
– callculus
Dec 7 '18 at 19:32
1
$begingroup$
Thank you, now I've understood my mistake
$endgroup$
– Martina Marty
Dec 7 '18 at 19:41
$begingroup$
That makes me happy. You´re welcome.
$endgroup$
– callculus
Dec 7 '18 at 19:43