How to prove that$D<0.5 T;$?
$begingroup$
We have the formula of obligation duration:
$$D = frac{C}{P} (1- frac{1+Tr}{(1+r)^{T}}) frac {1+r}{r^2} + frac {C+N}{P} frac{T}{(1+r)^2},$$
Here:
$rin(0,1)$,
$Tin mathbb N$ (number of time periods),
$Nin mathbb N$ (obligation nominal value),
$P in mathbb N$ (obligation price), $C=cN, cin(0,1).$
Check if $D<dfrac{T}{2}.$
sequences-and-series inequality
edited Jan 20 at 15:36
jordan_glen
1
asked Jan 20 at 15:33
PhilipPhilip
766
$endgroup$
add a comment |
$begingroup$
We have the formula of obligation duration:
$$D = frac{C}{P} (1- frac{1+Tr}{(1+r)^{T}}) frac {1+r}{r^2} + frac {C+N}{P} frac{T}{(1+r)^2},$$
Here:
$rin(0,1)$,
$Tin mathbb N$ (number of time periods),
$Nin mathbb N$ (obligation nominal value),
$P in mathbb N$ (obligation price), $C=cN, cin(0,1).$
Check if $D<dfrac{T}{2}.$
sequences-and-series inequality
edited Jan 20 at 15:36
jordan_glen
1
asked Jan 20 at 15:33
PhilipPhilip
766
$endgroup$
add a comment |
$begingroup$
We have the formula of obligation duration:
$$D = frac{C}{P} (1- frac{1+Tr}{(1+r)^{T}}) frac {1+r}{r^2} + frac {C+N}{P} frac{T}{(1+r)^2},$$
Here:
$rin(0,1)$,
$Tin mathbb N$ (number of time periods),
$Nin mathbb N$ (obligation nominal value),
$P in mathbb N$ (obligation price), $C=cN, cin(0,1).$
Check if $D<dfrac{T}{2}.$
sequences-and-series inequality
edited Jan 20 at 15:36
jordan_glen
1
asked Jan 20 at 15:33
PhilipPhilip
766
$endgroup$
We have the formula of obligation duration:
$$D = frac{C}{P} (1- frac{1+Tr}{(1+r)^{T}}) frac {1+r}{r^2} + frac {C+N}{P} frac{T}{(1+r)^2},$$
Here:
$rin(0,1)$,
$Tin mathbb N$ (number of time periods),
$Nin mathbb N$ (obligation nominal value),
$P in mathbb N$ (obligation price), $C=cN, cin(0,1).$
Check if $D<dfrac{T}{2}.$
sequences-and-series inequality
sequences-and-series inequality
edited Jan 20 at 15:36
jordan_glen
1
asked Jan 20 at 15:33
PhilipPhilip
766
edited Jan 20 at 15:36
jordan_glen
1
asked Jan 20 at 15:33
PhilipPhilip
766
edited Jan 20 at 15:36
jordan_glen
1
edited Jan 20 at 15:36
jordan_glen
1
edited Jan 20 at 15:36
jordan_glen
1
1
asked Jan 20 at 15:33
PhilipPhilip
766
asked Jan 20 at 15:33
PhilipPhilip
766
asked Jan 20 at 15:33
PhilipPhilip
766
766
add a comment |
add a comment |
1 Answer
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$begingroup$
Your assumption is false.
Counterexample: Define the following numbers
$$r = frac{1}{2}, $$
$$T = 1, $$
$$N = 1, $$
$$P = 1, $$
$$C = frac{1}{2}cdot 1 = frac{1}{2}. $$
Then
begin{align*}D &= frac{C}{P} left(1- frac{1+Tr}{(1+r)^{T}}right) frac {1+r}{r^2} + frac {C+N}{P} frac{T}{(1+r)^2}\
&= frac{1}{2}left(1 - frac{1 + 1/2}{1 + 1/2}right)cdot frac{1+ 1/2}{1/4} + frac{1/2 + 1}{1}cdot frac{1}{(1+ 1/2)^2}\
&=frac{1}{1+1/2} = frac{2}{3}.
end{align*}
Once $$D = frac{2}{3}> frac{1}{2} = frac{T}{2},$$ we concluded that $$D< frac{T}{2}$$ is, in general, false.
answered Jan 20 at 16:58
Matheus ManzattoMatheus Manzatto
1,3951524
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
votes
active
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active
oldest
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$begingroup$
Your assumption is false.
Counterexample: Define the following numbers
$$r = frac{1}{2}, $$
$$T = 1, $$
$$N = 1, $$
$$P = 1, $$
$$C = frac{1}{2}cdot 1 = frac{1}{2}. $$
Then
begin{align*}D &= frac{C}{P} left(1- frac{1+Tr}{(1+r)^{T}}right) frac {1+r}{r^2} + frac {C+N}{P} frac{T}{(1+r)^2}\
&= frac{1}{2}left(1 - frac{1 + 1/2}{1 + 1/2}right)cdot frac{1+ 1/2}{1/4} + frac{1/2 + 1}{1}cdot frac{1}{(1+ 1/2)^2}\
&=frac{1}{1+1/2} = frac{2}{3}.
end{align*}
Once $$D = frac{2}{3}> frac{1}{2} = frac{T}{2},$$ we concluded that $$D< frac{T}{2}$$ is, in general, false.
answered Jan 20 at 16:58
Matheus ManzattoMatheus Manzatto
1,3951524
$endgroup$
add a comment |
$begingroup$
Your assumption is false.
Counterexample: Define the following numbers
$$r = frac{1}{2}, $$
$$T = 1, $$
$$N = 1, $$
$$P = 1, $$
$$C = frac{1}{2}cdot 1 = frac{1}{2}. $$
Then
begin{align*}D &= frac{C}{P} left(1- frac{1+Tr}{(1+r)^{T}}right) frac {1+r}{r^2} + frac {C+N}{P} frac{T}{(1+r)^2}\
&= frac{1}{2}left(1 - frac{1 + 1/2}{1 + 1/2}right)cdot frac{1+ 1/2}{1/4} + frac{1/2 + 1}{1}cdot frac{1}{(1+ 1/2)^2}\
&=frac{1}{1+1/2} = frac{2}{3}.
end{align*}
Once $$D = frac{2}{3}> frac{1}{2} = frac{T}{2},$$ we concluded that $$D< frac{T}{2}$$ is, in general, false.
answered Jan 20 at 16:58
Matheus ManzattoMatheus Manzatto
1,3951524
$endgroup$
add a comment |
$begingroup$
Your assumption is false.
Counterexample: Define the following numbers
$$r = frac{1}{2}, $$
$$T = 1, $$
$$N = 1, $$
$$P = 1, $$
$$C = frac{1}{2}cdot 1 = frac{1}{2}. $$
Then
begin{align*}D &= frac{C}{P} left(1- frac{1+Tr}{(1+r)^{T}}right) frac {1+r}{r^2} + frac {C+N}{P} frac{T}{(1+r)^2}\
&= frac{1}{2}left(1 - frac{1 + 1/2}{1 + 1/2}right)cdot frac{1+ 1/2}{1/4} + frac{1/2 + 1}{1}cdot frac{1}{(1+ 1/2)^2}\
&=frac{1}{1+1/2} = frac{2}{3}.
end{align*}
Once $$D = frac{2}{3}> frac{1}{2} = frac{T}{2},$$ we concluded that $$D< frac{T}{2}$$ is, in general, false.
answered Jan 20 at 16:58
Matheus ManzattoMatheus Manzatto
1,3951524
$endgroup$
Your assumption is false.
Counterexample: Define the following numbers
$$r = frac{1}{2}, $$
$$T = 1, $$
$$N = 1, $$
$$P = 1, $$
$$C = frac{1}{2}cdot 1 = frac{1}{2}. $$
Then
begin{align*}D &= frac{C}{P} left(1- frac{1+Tr}{(1+r)^{T}}right) frac {1+r}{r^2} + frac {C+N}{P} frac{T}{(1+r)^2}\
&= frac{1}{2}left(1 - frac{1 + 1/2}{1 + 1/2}right)cdot frac{1+ 1/2}{1/4} + frac{1/2 + 1}{1}cdot frac{1}{(1+ 1/2)^2}\
&=frac{1}{1+1/2} = frac{2}{3}.
end{align*}
Once $$D = frac{2}{3}> frac{1}{2} = frac{T}{2},$$ we concluded that $$D< frac{T}{2}$$ is, in general, false.
answered Jan 20 at 16:58
Matheus ManzattoMatheus Manzatto
1,3951524
answered Jan 20 at 16:58
Matheus ManzattoMatheus Manzatto
1,3951524
answered Jan 20 at 16:58
Matheus ManzattoMatheus Manzatto
1,3951524
answered Jan 20 at 16:58
Matheus ManzattoMatheus Manzatto
1,3951524
1,3951524
add a comment |
add a comment |
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