We have the utility function.

$$U_{t} = \ln{c_{t}} + E_{t}\sum_{s=1}^{\infty}(\beta^{s}\ln{c_{t+s}})$$ And I am trying to find the value function.

$U$ is utility function. $c_t$ is consumption at time $t$. $\beta$ is the discount factor.

My manual says multiply the right side with $1-\beta$ to get:

$$U_{t} = (1-\beta)\ln c_{t} + \beta E_{t}U_{t+1}$$

How? And multiply only the right side? I am unsure what is happening here and how.

For $s = 1$ we have

$$U_{t} = \ln c_{t} + \beta E_{t}\ln c_{t+1}$$

but this is as far as I can understand. I don't get anywhere multiplying the equation by $(1-\beta)$.

  • $\begingroup$ I believe multiplying only the right side would be considered a positive monotonic transformation. So it would preserve the underlying preferences. This implies that the indifference curves are the same and the ordering of the 'bundles' are the same as before. I am not sure but I imagine this would imply that the value functions would be identical? That would be a good place to look into while waiting for someone to help you with this! $\endgroup$ – Brennan Oct 15 '19 at 21:15
  • $\begingroup$ Hi! Just to let you know you could use $\LaTeX$ to format equations in SE. $\endgroup$ – Art Oct 16 '19 at 2:24
  • 1
    $\begingroup$ @Art I provided an edit doing just that. Just awaiting OP approval $\endgroup$ – Brennan Oct 16 '19 at 2:32
  • $\begingroup$ I tried latex but nothing happened in the preview $\endgroup$ – MyJAJAJAJJA Oct 16 '19 at 14:47
  • $\begingroup$ But even if multiplying only one side, I can't understand how we end up here $\endgroup$ – MyJAJAJAJJA Oct 16 '19 at 14:47

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