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Asset Pricing John H. Cochrane June 12, 2000

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SECTION 1.5 DISCOUNT FACTORS IN CONTINUOUS TIME<br />

variables means that the finite time distribution of prices f(pt+∆|It) need not be normal.<br />

We can think of a riskfree security as one that has a constant price equal to one and pays<br />

the riskfree rate as a dividend,<br />

p =1; Dt = r f<br />

t , (25)<br />

or as a security that pays no dividend but whose price climbs deterministically at a rate<br />

dpt<br />

pt<br />

= r f<br />

t dt. (26)<br />

Next, we need to express the first order conditions in continuous time. The utility function<br />

is<br />

Z ∞<br />

U ({ct}) =E e<br />

t=0<br />

−δt u(ct)dt.<br />

Suppose the investor can buy a security whose price is pt and that pays a dividend stream Dt.<br />

As we did in deriving the present value price relation in discrete time, the first order condition<br />

for this problem gives us the infinite period version of the basic pricing equation right away1 ,<br />

ptu 0 (ct) =Et<br />

This equation is an obvious continuous time analogue to<br />

Z ∞<br />

e<br />

s=0<br />

−δs u 0 (ct+s)Dt+sds (27)<br />

∞X<br />

pt = Et β<br />

j=0<br />

t u0 (ct+j)<br />

u0 (ct) Dt+j.<br />

It turns out that dividing by u 0 (ct) is not a good idea in continuous time, since the ratio<br />

u 0 (ct+∆)/u 0 (ct) isn’t well behaved for small time intervals. Instead, we keep track of the<br />

level of marginal utility. Therefore, define the “discount factor” in continuous time as<br />

Then we can write the pricing equation as<br />

Λt ≡ e −δt u 0 (ct).<br />

Z ∞<br />

ptΛt = Et<br />

s=0<br />

Λt+sDt+sds. (28)<br />

1 One unit of the security pays the dividend stream Dt, i.e.Dtdt units of the numeraire consumption good in a<br />

time interval dt. The security costs pt units of the consumption good. The investor can finance the purchase of ξ<br />

units of the security by reducing consumption from et to ct = et − ξpt/dt during time interval dt. Thelossin<br />

utility from doing so is u 0 (ct)(et − ct)dt = u 0 (ct)ξpt. The gain is the right hand side of (1.27)<br />

35

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