Actuarially equivalent cashflow vector

Question

Suppose

$v(2,7)=0.5, v(2,8)=0.4,v(2,9)=0.3.$

Find a vector $\mathbf{d}$ actuarially equivalent to $\mathbf{c}=(1,3,5,2,9,10,6,4,8,3)$ such that $d_i=c_i$ for $i\lt7$ and $\mathbf{d}_i=0$ for $i\gt7$.

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For $i\gt7$,

$val_2(c,v)=C_7 v(2,7)+C_8v(2,8)+C_9v(2,9)$

$=(4)(0.5)+(8)(0.4)+(3)(0.3)=6.1$

I assume the new cashflow vector, $\mathbf{d}=[0_2,x,0_6]$

Equating for $x$

Since actuarially equivalent,

$6.1=xv(2,2) \rightarrow x=6.1$

Answer 1

From the information you have provided, I have made the following assumptions:

• $\mathbf{d}=(1,3,5,2,9,10,6,x,0,0)$
• $\mathbf{d}$ has a valuation function $v_\mathbf{d}(k)=1$, as we are not told anything to the contrary
• $v_\mathbf{c}(2,k)=1$ for $0\le k\le6$ for the cashflow vector $\mathbf{c}$

Actuarial equivalence between two cashflow vectors requires that:

$$\sum_\limits{k=0}^n \mathbf{c}[k]\cdot val_\mathbf{c}(k)=\sum_\limits{k=0}^n \mathbf{d}[k]\cdot val_\mathbf d(k)$$

As $\mathbf{c}=\mathbf{d}$ up to $k=7$, and the valuation functions are the same over this range, we can ignore these values.

Your sum to $6.1$ is correct, and so $\mathbf{d}=(1,3,5,2,9,10,6,6.1,0,0)$.

Answer 2

Since $c_i=d_i$ for $i<7$ and since $d_i=0$ for $i>7$, we get

$d=[1,3,5,2,9,10,6,x,0,0]$

Due to actuarial equivalence,

$x v(2,7)+ (0)v(2,8)+ (0)v(2,9)=6.1$

$x(0.5)=6.1 \rightarrow x=12.2$