I am reading Stephen Taylor's Asset Dynamics book and came across something I didn't fully understand.

For an ARCH process, the return series is modeled as $r_t = \mu_t + h_t^{1/2}z_t$ where is $z_t$ is $D(0,1)$ and may be normal/non-normal and $h_t$ is conditional variance (some function of subset of $\theta$).

Then he says the conditional density of $r_t$ is $f(r_t|\theta) = \frac{f(z_t(\theta)|\theta)}{\sqrt{h_t(\theta)}}$.

I don't understand how does the conditional variance term occur in the denominator of density relation and why doesn't $\mu_t$ occur in the relation. Can someone help me how to derive this?

  • $\begingroup$ What is $D(0,1)$? Space of càdlàg functions? $\endgroup$
    – mpiktas
    Mar 21, 2011 at 8:04

1 Answer 1


If the formula in the question is exactly how it is written in the book, then this is a bit of slack notation, with the ambiguous looking $f(.)$. The subscripts, while a bit ugly, are one way to make it more clear what the function represents (for $f(.)$ is essentially defined as two different things, which in "nit-picking" world is a contradiction unless $f(.)=0$ or $f(.)=\pm\infty$). In my notation,a capital letter stands for the "random variable" and a small letter for a potential value of that random variable. And the subscripts represent what the distribution is for (e.g. $f_{X|Y}(x|y)$ is the pdf for $X$, given $Y$ evaluated at values $x$ and $y$). I also used a capital $F$ for CDFs and small $f$ for PDFs.

Now I could have just written down "use the jacobian" as an answer, but I can never quite remember exactly what jacobian to use (is it J, or 1 over inverse J, or inverse J?). Going through the tedious process below (of essentially re-deriving the change-of-variables formula for PDFs) is the only way that I feel sure I've got the right answer.

AS far as I can tell, what has occurred here is that you define a "random variable" or "source of uncertainty", and call it $Z_t$. This has an unspecified PDF, denoted by $f_{Z_t}(z_{t})$. I am going to assume that by $D(0,1)$ you mean that the pdf is to have mean 0 and variance 1, but is otherwise ambiguous.

So we write the cumulative distribution function (CDF) for $R_t$ as:


Now $R_t$, given $\theta$ is a 1-to-1 function of $Z_t$. Now we substitute this relation $R_t=\mu_t(\theta)+\sqrt{h_t(\theta)}Z_t$ and re-arrange the expression to get:


Now you can see that this last expression is just the CDF of $Z_t$ evaluated at $\frac{r_t-\mu_t(\theta)}{\sqrt{h_t(\theta)}}$. I assume this is what is meant by $z_t(\theta)$. Now to get from the CDF to the PDF, you need to differentiate with respect to $r_t$.

$$f_{R_t|\theta}(r_t|\theta)=\frac{\partial}{\partial r_t}\left[F_{R_t|\theta}(r_t|\theta)\right]=\frac{\partial}{\partial r_t}\left[F_{Z_t|\theta}(z_t(\theta)|\theta)\right]$$

Now you use the chain rule for differentiation to get:

$$f_{R_t|\theta}(r_t|\theta)=\frac{\partial z_t(\theta)}{\partial r_t}\frac{\partial}{\partial z_t(\theta)}\left[F_{Z_t|\theta}(z_t(\theta)|\theta)\right]=\frac{\partial z_t(\theta)}{\partial r_t}f_{Z_t|\theta}(z_t(\theta)|\theta)$$

Now the derivative here can be calculated as:

$$\frac{\partial z_t(\theta)}{\partial r_t}=\frac{\partial \left[\frac{r_t-\mu_t(\theta)}{\sqrt{h_t(\theta)}}\right]}{\partial r_t}=\frac{1}{\sqrt{h_t(\theta)}}$$

Substituting this back into the answer gives the density required:


Note that this actually holds regardless of the distribution of the $Z_t$. For nowhere did I make use of the $D(0,1)$ assumption.


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