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Under the summation it says $\tau=t$. So $\tau$ take the value of $t$ in the first Beta, $t+1$ in the 2nd beta, $t+2$ in the 3rd beta and so on (Is this right?)

If the above statement is right, first beta term will be $\beta^{t-t} = 0$, 2nd beta term will be $\beta^{t+1-t} = \beta^1$, and $\beta^2$ so on.

Following this logic, $\beta$ increases in this fashion, $\beta^0,\,\beta^1, \,\beta^2$, which is exactly the same thing you would get if you put $t=0$ under the summation sign and Beta^t as the first term in the summation.

What am I missing here?

  • $k_\tau$ may depend on more than the difference between $\tau$ and $t$. – guest196883 Dec 07 '13 at 18:16
  • @SamDeHority that is it! I kept looking at Beta and I missed the K_tau which is what im really looking for. If you want to write that as an answer ill accept it –  Dec 07 '13 at 18:18

2 Answers2

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The first couple terms are

$$\beta^{0}u(F(k_t) - g_0(k_t)) + \beta^{1}u(F(k_{t+1}) - g_0(k_{t+1})) + $$ $$\beta^{2}u(F(k_{t+2}) - g_0(k_{t+2})) + \cdots$$

Does that clarify anything?

Elchanan Solomon
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Your given expression is: $$w_0(k_t) = \sum_{t = \tau}^\infty \beta^{t-\tau}u(F(k_\tau) - g_0(k_\tau))$$

$k_\tau$ depends on more than just the difference between $t$ and $\tau$, so do the summands, and therefore the sum.

guest196883
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