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I saw this in a paper (however, I cannot derive it): We define the growth rate of consumption as $g_{C, s} \equiv \frac{C_s}{C_{s-1}}$ and use $C_t=\left(\prod_{s=1}^t g_{C, s}\right)C_0$ to rewrite lifetime utility, $\mathcal{V}_0=$ $\mathbb{E}_0 \sum_{t=0}^{\infty} \beta^t \log \left(C_t\right)$, as $$ \mathcal{V}_0=\frac{1}{1-\beta}\left[\log \left(C_0\right)+\mathbb{E}_0\left(\sum_{t=1}^{\infty} \beta^t \log \left(g_{C, t}\right)\right)\right] $$

My attempt:

$$ \mathcal{V}_0= \mathbb{E}_0 \sum_{t=0}^{\infty} \beta^t \log \left(C_t\right) =\mathbb{E}_0 \sum_{t=0}^{\infty} \beta^t \log \left( \left(\prod_{s=1}^t g_{C, s}\right) C_0 \right) $$ $$ \mathcal{V}_0 = \mathbb{E}_0 \Big[ \frac{1}{1-\beta } \log C_0 + \sum_{t=0}^{\infty} \beta^t \log \left(\prod_{s=1}^t g_{C, s}\right) \Big] $$ $$ \mathcal{V}_0 = \mathbb{E}_0 \Big[ \frac{1}{1-\beta } \log C_0 + \sum_{t=0}^{\infty} \beta^t \left(\sum_{s=1}^t \log g_{C, s}\right) \Big] $$ $$ \mathcal{V}_0 = \Big[ \frac{1}{1-\beta } \log C_0 + \mathbb{E}_0 \left( \sum_{t=0}^{\infty} \beta^t \Big(\sum_{s=1}^t \log g_{C, s}\Big) \right) \Big] $$

Why is $\mathbb{E}_0 \left( \sum_{t=0}^{\infty} \beta^t \Big(\sum_{s=1}^t \log g_{C, s}\Big) \right)= \frac{1}{1-\beta } \mathbb{E}_0 \Big( \sum_{t=1}^{\infty} \beta^t \log g_{C, t} \Big)$?

Emmanuel Ameyaw
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1 Answers1

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More generally, $$ \sum_{t=0}^{\infty} \beta^t \Big(\sum_{s=1}^t \log g_{C, s}\Big)= \frac{1}{1-\beta } \sum_{t=1}^{\infty} \beta^t \log g_{C, t} . $$ Do it by interchange of the order of summation and summing a geometric series: \begin{align} \sum_{t=0}^{\infty} \beta^t \Big(\sum_{s=1}^t \log g_{C, s}\Big) &=\sum_{t=1}^{\infty} \beta^t \Big(\sum_{s=1}^t \log g_{C, s}\Big) \\&=\sum_{s=1}^\infty \Big(\sum_{t=s}^{\infty} \beta^t\log g_{C, s}\Big) \\&=\sum_{s=1}^\infty \Big(\sum_{t=s}^{\infty} \beta^t\Big)\log g_{C, s} \\&=\sum_{s=1}^\infty \Big(\frac{\beta^s}{1-\beta}\Big)\log g_{C, s} \\&=\frac{1}{1-\beta } \sum_{t=1}^{\infty} \beta^t \log g_{C, t} \end{align} Of course there are conditions that have to be checked to interchange the order of summation and to sum a geometric series. For example, it works if: $$ 0 < \beta<1, \quad g_{C,s} \ge 1 . $$

GEdgar
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