# How to get this budget constraint?

I have an Euler equation $(\frac{ c_ {t+1} }{c_t})^\sigma = \beta (1+r)$,where $c_t$ is the t period consumption,r is interest rate and $\beta$ is discount rate, and a budget constraint $\sum _{t=0}^\infty \frac{c_t}{(1+r)^t}=\sum _{t=0}^\infty \frac{y_t}{(1+r)^t}$, where $y_t$ is income.

My question is how to substitute Euler equation into budget constraint to get the following equation?

$c_t\sum_{j=0}^\infty\frac{[\beta(1+r)]^\frac{j}{\sigma}}{(1+r)^j}=\sum _{t=0}^\infty \frac{y_t}{(1+r)^t}$

The hint is to use the recursive Euler equation to express all future consumptions $c_{t+j}$ as a function of current consumption $c_t$, and to play with the indexes in the sum of the left.

• This would be more appropriate as a comment than an answer. – Kitsune Cavalry Jun 29 '16 at 18:00

Hint: Separate the consumption time periods \begin{align} \left(\frac{ c_ {t+1} }{c_t}\right)^\sigma & = \beta (1+r) \\ \frac{ c_ {t+1} }{c_t} & = \left[\beta (1+r) \right]^\frac{1}{\sigma} \\ c_ {t+1} & = c_t \left[\beta (1+r) \right]^\frac{1}{\sigma} \\ \end{align}

Notice that $(\frac{ c_ {t+2} }{c_{t+1}})^\sigma = \beta (1+r)$. So,

$$c_ {t+2} = c_{t+1} \left[\beta (1+r) \right]^\frac{1}{\sigma} = c_{t} \left[\beta (1+r) \right]^{\frac{1}{\sigma} \cdot 2}$$

So

$$\sum _{t=0}^\infty \frac{c_t}{(1+r)^t}$$

can be expressed as a geometric series and simplified. Working that out for yourself should give you the desired answer.

• How did you get the 3rd equation from the 2nd one? – XJ.C Jun 30 '16 at 2:39
• I made a mistake. I'll amend the solution. – Kitsune Cavalry Jun 30 '16 at 3:16
• Following your hint, I can get $\sum _{t=0}^\infty \frac{c_t}{(1+r)^t}=c_t\sum_{j=0}^\infty\frac{[\beta(1+r)]^\frac{j}{\sigma}}{(1+r)^j}$. This is calculate the PV of consumption flow. LHS start from period 0 to $\infty$,while RHS,$\sum _{j=0}^\infty \frac{c_{t+j}}{(1+r)^j}$, from period t to $\infty$. I am confuse about this. – XJ.C Jun 30 '16 at 23:39
• I don't think there's any meaningful distinction between $j$ and $t$. Just different ways to denote time. – Kitsune Cavalry Jul 1 '16 at 0:03
• I mean that starting from period 0 and from t (Say t=9). – XJ.C Jul 1 '16 at 0:25