Suppose we have a stationary AR(1) process: $Y_{t+1}=a+ \rho Y_{t} + \epsilon_{t+1}$ where $\epsilon_{t+1}$ is white noise with probability density function $\phi(.)$.

Now say we have a equation $P_{t+1}=u\cdot (A_{m}-Y_{t+1})^{+}$ where $(Y_{m}-Y_{t+1})^{+}$ = $ \max \{(A_{m}-Y_{t+1}),0\}$ and $u$ is a constant.

So $P_{t+1}=u\cdot(A_{m}-a-\rho Y_{t}-\epsilon_{t+1})^{+}$.

Can we write $E[P_{t+1} \mid Y_{t}]= u\cdot \int_{-\infty}^{(A_{m}-a - \rho Y_{t})}(A_{m}-a-\rho Y_{t}-\epsilon_{t+1})\phi(\epsilon_{t+1})d\epsilon_{t+1}$ ?

  • $\begingroup$ Are you assuming distribution of $\epsilon_t$ identical? Also what about $m$? $m<t$? $\endgroup$
    – vinux
    Feb 8, 2014 at 7:28
  • $\begingroup$ Yes $\epsilon_{t}$ i.i.d.; Also $A_{m}$ is a constant. $\endgroup$
    – Pradipta
    Feb 8, 2014 at 7:45
  • $\begingroup$ Sorry $Y_{m}$ is a constant. $\endgroup$
    – Pradipta
    Feb 8, 2014 at 8:47

1 Answer 1


Answer is Yes.

Let $X$ is a continuous random variable with pdf $f(x)$.

Then $$ \begin{align*} \operatorname {E}[(k-X)^{+}] & = \int_{-\infty}^{\infty}(k-x)^{+} f(x)dx \\ &= \int_{-\infty}^{k}(k-x)^{+} f(x)dx + \int_{k}^{\infty}(k-x)^{+} f(x)dx \\ &= \int_{-\infty}^{k}(k-x) f(x)dx + 0= \int_{-\infty}^{k}(k-x) f(x)dx \end{align*} $$ Since $(k-x)^{+}$ is $(k-x)$ when $x<a$ and 0 otherwise.

Given $Y_t$, $Y_{m}-a-\rho Y_{t}$ is a constant and you can replace it as $k$. If $\epsilon_t$ is a continuous random variable, you can use the above result and expression will be same as you wrote.


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