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I want to solve the integral

$$I =\int_{0}^{3} \frac{\exp(-s)}{1+\frac {1}{s}} \text{d}s $$

using importance sampling.

I'm unsure as to how implement it.

I have sampled random variables from an exponential distribution.

$$ x = -\log(1-U)\ \text{with}\ U \sim \mathcal{U}(0,1) $$

Am I right in thinking that

$$ \hat{I} = \frac{1}{N}\sum f(x)/g(x) $$

is a proper importance sampling approximation, where $$f(s)=\frac{\exp(-s)}{1+\frac {1}{s}}$$

and $g$ is the pdf of the standard exponential distribution.

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1 Answer 1

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Pretty close. Remember that exponentials have support on the positive reals.

$$ \frac{1}{N}\sum_i f(x_i)/g(x_i) \to E_g[f(x)/g(x)] = \int_0^{\infty}f(x)/g(x) g(x)dx = \int_0^{\infty}f(x) dx \neq \int_0^3 f(x) dx. $$

Instead try computing $\frac{1}{N}\sum_i f(x_i)/g(x_i)1(0 < x_i < 3)$.

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  • $\begingroup$ Ok great, that makes sense. $\endgroup$ Commented Dec 3, 2017 at 12:16

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