0
$\begingroup$

I am reading chapter 11 . Sampling Methods from the book : Pattern Recognition and Machine Learning by Bishop : In the introduction , in short,he evaluates expectation of some function $f(z)$ with respect to a probability distribution $P(z)$ where $z$ is the random variable . He writes :
\begin{align} E(f) = \int f(z)p(z)dz \end{align}
Now he says that we suppose that such expectations are too complex to be evaluated exactly using analytical techniques . So we use sampling. And the idea behind sampling methods is to obtain a set of samples $z^{(l)}$, (where $l=1,....,L$) drawn independently from the distribution $P(z)$. This allows the expectation to be written as \begin{align} f^{\hat{}} = \frac{\sum_{l=1}^{L}f(z^{(l)})}{L} \end{align} Now further in the official solution for excercise 11.1 , he calculates : \begin{align} E[f^{\hat{}}] = \frac{1}{L}\sum_{l=1}^{L}\int f(z^{(l)}) p(z^{(l)})dz^{(l)}\tag 1 \end{align} Now i did not understand this integral .
My argument , for simplicity , let us assume the underlying distribution $P(z)$ to be a one dimensional standard normal distribtion .
1.) When the author samples $z^{1}$, i assume that he drew a fixed $n$ values which would look something like : $z^{1}= \{0.2,0.30.2323,... ,0.8\}$ ,total $n$ values .
Now in that case how would the integeral in $(1)$ look like ? For example what would be the limits of integration ?

$\endgroup$
0
$\begingroup$

If $P(z)$ is a one-dimensional normal distribution, then there is no $n$. Each sample $z^{(l)}$ is a single draw from $P(z)$, i.e. $z^{(l)}$ is just a real number. The integral is over the support of the distribution. In this case, the limits of integration are $(-\infty, \infty)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.