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When to sell a coin collection
Expected Ratio of Coin FlipsHelp understanding confusing present value question (Theory of Interest)Coin-tossing gamesCoin Flipping Game - Expected Number of TailsA coin is tossed three times. Given that at least one head appears, what is the probability that exactly two will appear?A coin is tossed if a dice is rolledCoin flipping problemConfounding Coin Chuck ChancesA fair coin tossed repeatedlyProbability of Die and Coin
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Problem: A person inherits a coin collection that has a value of $2000e^sqrtt$ m.u. (monetary unit) after $t$ years. Assume a bank interest of 7% per year. At which time $t_0$ years (if there is one) is it best to sell the coin collection and deposit the money at the bank? (The life expectancy of the owner is disregarded and the collection is passed on to any surviving relatives, so $t$ is not limited by the remaining years of the current owner.)
Solution: The definition of ”best time” is unclear to me. I thought of it as follows: Let $V(t)=2000e^sqrtt$ be the value of the coin collection at time $t$ years. The increased value of the collection per year should be less than the annual 7% interest of the collection’s value at $t$, when it is sold, i.e.
$$
V(t+1)-V(t)<0.07V(t)
$$
which has a ”break point” at $t=54$ years, but the answer is 57 years. Any thoughts what could be wrong?
problem-solving
$endgroup$
add a comment |
$begingroup$
Problem: A person inherits a coin collection that has a value of $2000e^sqrtt$ m.u. (monetary unit) after $t$ years. Assume a bank interest of 7% per year. At which time $t_0$ years (if there is one) is it best to sell the coin collection and deposit the money at the bank? (The life expectancy of the owner is disregarded and the collection is passed on to any surviving relatives, so $t$ is not limited by the remaining years of the current owner.)
Solution: The definition of ”best time” is unclear to me. I thought of it as follows: Let $V(t)=2000e^sqrtt$ be the value of the coin collection at time $t$ years. The increased value of the collection per year should be less than the annual 7% interest of the collection’s value at $t$, when it is sold, i.e.
$$
V(t+1)-V(t)<0.07V(t)
$$
which has a ”break point” at $t=54$ years, but the answer is 57 years. Any thoughts what could be wrong?
problem-solving
$endgroup$
$begingroup$
I collect coins so I'm interested to see the answer to this.
$endgroup$
– Gnumbertester
8 hours ago
add a comment |
$begingroup$
Problem: A person inherits a coin collection that has a value of $2000e^sqrtt$ m.u. (monetary unit) after $t$ years. Assume a bank interest of 7% per year. At which time $t_0$ years (if there is one) is it best to sell the coin collection and deposit the money at the bank? (The life expectancy of the owner is disregarded and the collection is passed on to any surviving relatives, so $t$ is not limited by the remaining years of the current owner.)
Solution: The definition of ”best time” is unclear to me. I thought of it as follows: Let $V(t)=2000e^sqrtt$ be the value of the coin collection at time $t$ years. The increased value of the collection per year should be less than the annual 7% interest of the collection’s value at $t$, when it is sold, i.e.
$$
V(t+1)-V(t)<0.07V(t)
$$
which has a ”break point” at $t=54$ years, but the answer is 57 years. Any thoughts what could be wrong?
problem-solving
$endgroup$
Problem: A person inherits a coin collection that has a value of $2000e^sqrtt$ m.u. (monetary unit) after $t$ years. Assume a bank interest of 7% per year. At which time $t_0$ years (if there is one) is it best to sell the coin collection and deposit the money at the bank? (The life expectancy of the owner is disregarded and the collection is passed on to any surviving relatives, so $t$ is not limited by the remaining years of the current owner.)
Solution: The definition of ”best time” is unclear to me. I thought of it as follows: Let $V(t)=2000e^sqrtt$ be the value of the coin collection at time $t$ years. The increased value of the collection per year should be less than the annual 7% interest of the collection’s value at $t$, when it is sold, i.e.
$$
V(t+1)-V(t)<0.07V(t)
$$
which has a ”break point” at $t=54$ years, but the answer is 57 years. Any thoughts what could be wrong?
problem-solving
problem-solving
edited 8 hours ago
mf67
asked 8 hours ago
mf67mf67
354 bronze badges
354 bronze badges
$begingroup$
I collect coins so I'm interested to see the answer to this.
$endgroup$
– Gnumbertester
8 hours ago
add a comment |
$begingroup$
I collect coins so I'm interested to see the answer to this.
$endgroup$
– Gnumbertester
8 hours ago
$begingroup$
I collect coins so I'm interested to see the answer to this.
$endgroup$
– Gnumbertester
8 hours ago
$begingroup$
I collect coins so I'm interested to see the answer to this.
$endgroup$
– Gnumbertester
8 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Consider the money you'd have at the time $T$, assuming you sold the coin collection at the time $t<T$ and immediately deposit the money in the bank. You'd have $$ W(T,t) = V(t)cdot 1.07^T-t = 2000 e^sqrtt + (T-t)ln (1.07) $$
Considering $T$ to be constant, we want to find the best time of selling $t$. That means finding the maximum of function $$ f(t) = sqrtt + (T-t)ln(1.07)$$
We have $$ f'(t) = frac12sqrtt - ln(1.07)$$
$$ f'(t_0) = 0 qquad Leftrightarrow qquad t_0= frac14(ln(1.07))^2 approx 54.6127 $$
It does not depend on $T$, which means that no matter what $Tge t_0$ you consider, if you want to have the most money in the and, it's best to sell the coin collection at time $t_0$.
The answer 57 years is wrong.
$endgroup$
add a comment |
$begingroup$
The "force of interest" is the rate of growth divided by the value of the asset, so if the value at time $t$ is $f(t)$, the force of interest is $$f'(t)over f(t),$$ the logarithmic derivative of $f$. Therefore we want to sell when $$1over2sqrtt=log1.07$$ which happens when $t=54.6127.$ If you have to sell after a whole number of years, you'd have to work out whether it's best to sell after $54$ or $55$ years, but in any event, $57$ seems incorrect.
$endgroup$
add a comment |
$begingroup$
I would say your approach looks more or less correct. I would probably simplify it a bit in saying
$frac frac dVdtV = 0.07$
Well, it is simpler if you know calculus.
$frac dvdt = frac 2000e^sqrt t2sqrt t = frac V2sqrt t\
frac12sqrt t = 0.07$
Using the approach in the OP, you should be able to get there with a few guesses in a guess and check approach.
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the money you'd have at the time $T$, assuming you sold the coin collection at the time $t<T$ and immediately deposit the money in the bank. You'd have $$ W(T,t) = V(t)cdot 1.07^T-t = 2000 e^sqrtt + (T-t)ln (1.07) $$
Considering $T$ to be constant, we want to find the best time of selling $t$. That means finding the maximum of function $$ f(t) = sqrtt + (T-t)ln(1.07)$$
We have $$ f'(t) = frac12sqrtt - ln(1.07)$$
$$ f'(t_0) = 0 qquad Leftrightarrow qquad t_0= frac14(ln(1.07))^2 approx 54.6127 $$
It does not depend on $T$, which means that no matter what $Tge t_0$ you consider, if you want to have the most money in the and, it's best to sell the coin collection at time $t_0$.
The answer 57 years is wrong.
$endgroup$
add a comment |
$begingroup$
Consider the money you'd have at the time $T$, assuming you sold the coin collection at the time $t<T$ and immediately deposit the money in the bank. You'd have $$ W(T,t) = V(t)cdot 1.07^T-t = 2000 e^sqrtt + (T-t)ln (1.07) $$
Considering $T$ to be constant, we want to find the best time of selling $t$. That means finding the maximum of function $$ f(t) = sqrtt + (T-t)ln(1.07)$$
We have $$ f'(t) = frac12sqrtt - ln(1.07)$$
$$ f'(t_0) = 0 qquad Leftrightarrow qquad t_0= frac14(ln(1.07))^2 approx 54.6127 $$
It does not depend on $T$, which means that no matter what $Tge t_0$ you consider, if you want to have the most money in the and, it's best to sell the coin collection at time $t_0$.
The answer 57 years is wrong.
$endgroup$
add a comment |
$begingroup$
Consider the money you'd have at the time $T$, assuming you sold the coin collection at the time $t<T$ and immediately deposit the money in the bank. You'd have $$ W(T,t) = V(t)cdot 1.07^T-t = 2000 e^sqrtt + (T-t)ln (1.07) $$
Considering $T$ to be constant, we want to find the best time of selling $t$. That means finding the maximum of function $$ f(t) = sqrtt + (T-t)ln(1.07)$$
We have $$ f'(t) = frac12sqrtt - ln(1.07)$$
$$ f'(t_0) = 0 qquad Leftrightarrow qquad t_0= frac14(ln(1.07))^2 approx 54.6127 $$
It does not depend on $T$, which means that no matter what $Tge t_0$ you consider, if you want to have the most money in the and, it's best to sell the coin collection at time $t_0$.
The answer 57 years is wrong.
$endgroup$
Consider the money you'd have at the time $T$, assuming you sold the coin collection at the time $t<T$ and immediately deposit the money in the bank. You'd have $$ W(T,t) = V(t)cdot 1.07^T-t = 2000 e^sqrtt + (T-t)ln (1.07) $$
Considering $T$ to be constant, we want to find the best time of selling $t$. That means finding the maximum of function $$ f(t) = sqrtt + (T-t)ln(1.07)$$
We have $$ f'(t) = frac12sqrtt - ln(1.07)$$
$$ f'(t_0) = 0 qquad Leftrightarrow qquad t_0= frac14(ln(1.07))^2 approx 54.6127 $$
It does not depend on $T$, which means that no matter what $Tge t_0$ you consider, if you want to have the most money in the and, it's best to sell the coin collection at time $t_0$.
The answer 57 years is wrong.
answered 7 hours ago
Adam LatosińskiAdam Latosiński
6,9947 silver badges20 bronze badges
6,9947 silver badges20 bronze badges
add a comment |
add a comment |
$begingroup$
The "force of interest" is the rate of growth divided by the value of the asset, so if the value at time $t$ is $f(t)$, the force of interest is $$f'(t)over f(t),$$ the logarithmic derivative of $f$. Therefore we want to sell when $$1over2sqrtt=log1.07$$ which happens when $t=54.6127.$ If you have to sell after a whole number of years, you'd have to work out whether it's best to sell after $54$ or $55$ years, but in any event, $57$ seems incorrect.
$endgroup$
add a comment |
$begingroup$
The "force of interest" is the rate of growth divided by the value of the asset, so if the value at time $t$ is $f(t)$, the force of interest is $$f'(t)over f(t),$$ the logarithmic derivative of $f$. Therefore we want to sell when $$1over2sqrtt=log1.07$$ which happens when $t=54.6127.$ If you have to sell after a whole number of years, you'd have to work out whether it's best to sell after $54$ or $55$ years, but in any event, $57$ seems incorrect.
$endgroup$
add a comment |
$begingroup$
The "force of interest" is the rate of growth divided by the value of the asset, so if the value at time $t$ is $f(t)$, the force of interest is $$f'(t)over f(t),$$ the logarithmic derivative of $f$. Therefore we want to sell when $$1over2sqrtt=log1.07$$ which happens when $t=54.6127.$ If you have to sell after a whole number of years, you'd have to work out whether it's best to sell after $54$ or $55$ years, but in any event, $57$ seems incorrect.
$endgroup$
The "force of interest" is the rate of growth divided by the value of the asset, so if the value at time $t$ is $f(t)$, the force of interest is $$f'(t)over f(t),$$ the logarithmic derivative of $f$. Therefore we want to sell when $$1over2sqrtt=log1.07$$ which happens when $t=54.6127.$ If you have to sell after a whole number of years, you'd have to work out whether it's best to sell after $54$ or $55$ years, but in any event, $57$ seems incorrect.
answered 7 hours ago
saulspatzsaulspatz
22.4k4 gold badges16 silver badges38 bronze badges
22.4k4 gold badges16 silver badges38 bronze badges
add a comment |
add a comment |
$begingroup$
I would say your approach looks more or less correct. I would probably simplify it a bit in saying
$frac frac dVdtV = 0.07$
Well, it is simpler if you know calculus.
$frac dvdt = frac 2000e^sqrt t2sqrt t = frac V2sqrt t\
frac12sqrt t = 0.07$
Using the approach in the OP, you should be able to get there with a few guesses in a guess and check approach.
$endgroup$
add a comment |
$begingroup$
I would say your approach looks more or less correct. I would probably simplify it a bit in saying
$frac frac dVdtV = 0.07$
Well, it is simpler if you know calculus.
$frac dvdt = frac 2000e^sqrt t2sqrt t = frac V2sqrt t\
frac12sqrt t = 0.07$
Using the approach in the OP, you should be able to get there with a few guesses in a guess and check approach.
$endgroup$
add a comment |
$begingroup$
I would say your approach looks more or less correct. I would probably simplify it a bit in saying
$frac frac dVdtV = 0.07$
Well, it is simpler if you know calculus.
$frac dvdt = frac 2000e^sqrt t2sqrt t = frac V2sqrt t\
frac12sqrt t = 0.07$
Using the approach in the OP, you should be able to get there with a few guesses in a guess and check approach.
$endgroup$
I would say your approach looks more or less correct. I would probably simplify it a bit in saying
$frac frac dVdtV = 0.07$
Well, it is simpler if you know calculus.
$frac dvdt = frac 2000e^sqrt t2sqrt t = frac V2sqrt t\
frac12sqrt t = 0.07$
Using the approach in the OP, you should be able to get there with a few guesses in a guess and check approach.
answered 7 hours ago
Doug MDoug M
3,0691 gold badge5 silver badges12 bronze badges
3,0691 gold badge5 silver badges12 bronze badges
add a comment |
add a comment |
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$begingroup$
I collect coins so I'm interested to see the answer to this.
$endgroup$
– Gnumbertester
8 hours ago