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You are given $q_x=0.053$, $q_{x+1}=0.054$, $q_{x+2}=0.055$, $i_0=0.06$, $i_1=0.08$, $i_2=0.10$. If $\mathbf{b}=(1,2,3)$, find the vector $\mathbf{c=b*w_x}$

$(w_x)_1=v(0,1) q_x=(1.06)^{-1}(0.053)=0.05$

$(w_x)_2=v(1,2) q_{x+1}=(1.06)^{-1}(1.08)^{-1}(0.054)=0.0471$

$(w_x)_2=v(2,3) q_{x+2}=(1.06)^{-1}(1.08)^{-1}(1.10)^{-1}(0.055)=0.0436$

$\mathbf{c=b*w_x}=(1,2,3)*(0.05, 0.0471,0.0436)=(0.05,0.942,0..1308)$

But the answer is $(0.05, 0.1, 0.15)$

1 Answers 1

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I think your mistake is in the discount factors $v(t,t+1)$. You should have \begin{align*} v(1,2) &= 1.08^{-1}\\ v(2,3) &= 1.1^{-1}. \end{align*} If you re-do the calculations using these definitions, you should get the right answer.

Note that in general, $v(s,t)$ denotes the discount factor from time $t$ back to time $s$. By writing $1.06^{-1}1.08^{-1}$, what you get is $v(0,2)$ (instead of $v(1,2)$, which is what you want).