A discount function satisfies
$v(k)=2^{-k}[1-\frac{k}{6}], k=0,1,2,3,4,5.$
For the vector $\mathbf{c}=(1,-2,4,3,-3,-5).$ Find
(a) $_3V(\mathbf{c})$
(b) $B_3(\mathbf{c})$
Calculate discount function at each time
$v(0)=1, v(1)=\frac{5}{12}, v(2)=\frac{1}{6}, v(3)=\frac{1}{16}, v(4)=\frac{1}{48},v(5)=\frac{1}{192}$
I tried (b)
$B_3(\mathbf{c})=val(3)$
=$(1)(1)+(-2)(\frac{5}{12})+(4)(\frac{1}{6})$
=$\frac{5}{6}$
For (a)
$_3V(c)=val(8)-val(3)$
$=(3)(\frac{1}{16})+(-3)(\frac{1}{48})+(-5)(\frac{1}{192})$
$=\frac{19}{192}$
But I am not sure of my work.
Alternative method,
I get the balance at time 3 to be given as
$B_3(c)=C_0v(3.0)+C_1v(3,1)+C_2v(3,2)+...+C_5v(3,5)$
$=\frac{1}{v(3)}[C_Ov(0)+C_1v(1)+...+C_5v(5)]$
$=16[(1)(1)+(-2)(\frac{5}{12})+(4)(\frac{1}{6})+(3)(\frac{1}{6})+(-3)(\frac{1}{48})+(-5)(\frac{1}{192})]=14.916$
Then, I do not know $_3V(c)$
The balance at time 3 is given by
$$B_3(c) = C_0\frac{1}{v_3} + C_1\frac{v_1}{v_3} + C_2\frac{v_2}{v_3}+C_3$$
$$_3V(c)= B_3(c) + C_4 \frac{v_4}{v_3} + C_5\frac{v_5}{v_3}$$
Thus
$$B_3(c) = 16.33$$
$$_3V(c)= 14.917$$
Could you verify the answer, Please