Question

Let \(U\) have the uniform distribution on the interval\([0,1]\). Show that the random variable \(W\)defined in Eq. (12.4.6) has the p.d.f. \(h\)defined in Eq. (12.4.5).

Short answer

This is true because the p.d.f. of random variable with uniform distribution on \([0,1]\) is equal to\(l\).

Step-by-step solution

Definition for importance sampling

• Many integrals can be advantageously recast as random variable functions.
• We can estimate integrals that would otherwise be impossible to compute in closed form if we can simulate a large number of random variables with proper distributions.

Determine the inverse function and its derivative

Random variable \(W\)is defined as

\(W = {\mu _2} + {\sigma _2}{\Phi ^{ - 1}}\left[ {U\Phi \left( {\frac{{{c_2} - {\mu _2}}}{{{\sigma _2}}}} \right)} \right]\)And function \(h\) is defined as

\(h\left( {{x_2}} \right) = \frac{{{{\left( {2\pi \sigma _2^2} \right)}^{ - 1/2}}\exp \left[ {{{\left( {{x_2} - {\mu _2}} \right)}^2}/\left( {2\sigma _2^2} \right)} \right]}}{{\Phi \left[ {\left( {{c_2} - {\mu _2}} \right)/{\sigma _2}} \right]}},\;\;\; - \infty < {x_2} \le {c_2}\)

From

\(w = {\mu _2} + {\sigma _2}{\Phi ^{ - 1}}\left[ {u\Phi \left( {\frac{{{c_2} - {\mu _2}}}{{{\sigma _2}}}} \right)} \right]\)

It follows that

\(u = \frac{{\Phi \left[ {\left( {w - {\mu _2}} \right)/{\sigma _2}} \right]}}{{\Phi \left[ {\left( {{c_2} - {\mu _2}} \right)/{\sigma _2}} \right]}}\)Is the inverse transformation.

The derivative of \(\Phi \)is the p.d.f. of a standard normal distribution, hence, the derivative of the inverse function is

\(\frac{\partial }{{\partial w}}u = \frac{1}{{\Phi \left[ {\left( {{c_2} - {\mu _2}} \right)/{\sigma _2}} \right]}} \cdot {\left( {2\pi \sigma _2^2} \right)^{ - 1/2}}\exp \left[ {\frac{{{{\left( {{x_2} - {\mu _2}} \right)}^2}}}{{2\sigma _2^2}}} \right] = h\left( {{x_2}} \right)\)

Which is p.d.f. ofW.

This is true because the p.d.f. of random variable with uniform distribution on \([0,1]\) is equal to \(1.\)