Consider these two cases:

  • Case 1: $T_1\sim\operatorname{exp}(\lambda_1)$, $T_2\sim\operatorname{exp}(\lambda_2),T_3\sim\operatorname{exp}(\lambda_3)$,
  • Case 2: $T_1\sim\operatorname{exp}(\theta_1)$, $T_2\sim\operatorname{exp}(\theta_2),T_3\sim\operatorname{exp}(\theta_3)$.

We define $T_4 = \max(T_1 + T_2, T_3)$. Suppose that we want to estimate $\mu_2 = E[T_4]$, where $T_4$ is distributed according to case $2$. How can we do so using importance sampling based on a sample $T_1, \ldots, T_n$ generated from case $1$?

I want something like $$\mu_2 \approx \dfrac{1}{n}\sum\limits_{i = 1}^nT_{4i}\dfrac{f(T_{4i})}{q(T_{4i})}$$ where the $T_{4i}$ are generated based on the distributions from case $1$ and where $f$ is the density function of $T_4$ for case $2$ and $q$ is the density function of $T_4$ for case $1$.

Question: I know how to generate $T_{4i}$ based on case $1$, but I don't know how to figure out $f$ and $g$. How should I proceed?

Edit: As pointed out in the comments; I know that I can just use regular Monte Carlo to estimate $\mu_2$. I'm just interested in how I would use importance sampling if I wouldn't be able to simulate directly from case 2.

  • $\begingroup$ you can just use regular Monte Carlo and calculate $n^{-1}\sum_i T_{4i}$. Importance sampling would be useful if you couldn't generate from the target, but could only evaluate the density (perhaps only up to a normalizing constant). If this was for a homework assignment, and you needed to use importance sampling, you would have to derive the distribution for each $T_4$ $\endgroup$
    – Taylor
    Commented Apr 9, 2018 at 11:49
  • $\begingroup$ @Taylor Thank you for your reply! Yes I'm aware of that and I have done so already. I'm just interested in what this would look like if I wouldn't have been able to so and had to use importance sampling. $\endgroup$ Commented Apr 9, 2018 at 12:59
  • $\begingroup$ are you having trouble deriving $f$? Or choosing $q$? $\endgroup$
    – Taylor
    Commented Apr 9, 2018 at 14:03
  • $\begingroup$ @Taylor well both really! $\endgroup$ Commented Apr 9, 2018 at 15:17

1 Answer 1


Importance sampling is based on the identity $$\mathbb{E}_f[h(X)]=\int h(x) f(x)\text{d}x =\int h(x) \frac{f(x)}{g(x)}g(x)\text{d}x $$where both $f$ and $g$ are probability densities. This means one can use a sample generated from $g$ to approximate an expectation under $f$.

In the current problem, the density of$$T_4=\max\{T_1+T_2,T_3\}$$is not provided and while available within a few steps, it does not need to be derived. Another approach is indeed to consider $T_4$ as a function of $(T_1,T_2,T_3)$ and integrate under the distribution of this triplet: \begin{align*} \mathbb{E}_\theta[T_4]&=\mathbb{E}_\theta[\max\{T_1+T_2,T_3\}]\\ &=\int \max\{t_1+t_2,t_3\} f_\theta(t_1,t_2,t_3)\,\text{d}t_1\text{d}t_2\text{d}t_3\\ &=\int \max\{t_1+t_2,t_3\} \dfrac{f_\theta(t_1,t_2,t_3)}{f_\lambda(t_1,t_2,t_3)}f_\lambda(t_1,t_2,t_3)\,\text{d}t_1\text{d}t_2\text{d}t_3\\ &=\mathbb{E}_\lambda\left[T_4 \dfrac{f_\theta(T_1,T_2,T_3)}{f_\lambda(T_1,T_2,T_3)}\right] \end{align*}

  • 1
    $\begingroup$ "[I]t does not need to be derived" missed that. (+1) $\endgroup$
    – Taylor
    Commented Apr 11, 2018 at 16:00
  • $\begingroup$ Thank you for your answer. I'm interested in how I would use importance sampling here though. Could you show me how I would do so? $\endgroup$ Commented Apr 24, 2018 at 7:49
  • $\begingroup$ Which part of my answer is unclear? Importance sampling based on any importance function on $(T_1,T_2,T_3)$ can be used for estimating the expectation of $T_4$ or of any function of $T_4$. $\endgroup$
    – Xi'an
    Commented Apr 25, 2018 at 10:13

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