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I have a task in a subject called "Monte-Carlo methods" with which I'm a bit stuck and therefore I'm asking for your help.

The task is as follows: Describe in detail one specific option how to find approximately a value of the integral using Monte-Carlo methods and using ONLY independent random variables from uniform distribution with parametres (0,1):

$$\int_{2}^{4}(\int_{1}^{\infty} e^{x+y-xy} dx) dy$$

What I've thought so far is that Importance Sampling Method might be a good idea to solve it. So I would choose a suitable density functions for X ja Y. For X it would be a uniform distrbution with parametres (2,4) since it's values are bounded and for Y I would choose exponential distrbution with rate 1 (Exp(1)).

But the main problem relies in the second part of the question - how to use ONLY values from uniform distribution (0,1). I know that I have to use the inverse of a density function, but I'm not able to write it down as requested.

I hope you understand my question and can help me with this one!

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    $\begingroup$ I suspect the (easily demonstrated) fact that the log of a uniformly distributed random variable has an exponential distribution might be very useful for this problem. $\endgroup$ – whuber Jun 16 '18 at 12:07
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Hint #1: \begin{align*}\int_{2}^{4}\int_{1}^{\infty} e^{x+y-xy} \text{d}x \text{d}y&=\int_{2}^{4}\int_{0}^{1} e^{y+z^{-1}-z^{-1}y} \, z^{-2}\text{d}z\text{d}y\\&=\int_{0}^{1}\int_{0}^{1} e^{2(w+1)+z^{-1}-2z^{-1}(w+1)} \, z^{-2}\text{d}z2\text{d}w\end{align*} Hint #2: \begin{align*}\int_{2}^{4}\int_{1}^{\infty} e^{x+y-xy} \text{d}x \text{d}y&=\int_{2}^{4}y^{-1}e^{y}\,\left\{\int_{0}^{\infty} e^{x}\,ye^{-xy}\text{d}x\right\} \text{d}y\\ &=\int_{0}^{1} (2[w+1])^{-1}\exp\{2[w+1]\}\underbrace{\mathbb{E}^X[\exp(X)|Y=y]}_{X\sim\cal{E}xp(y)}2\text{d}w\\ &=\int_{0}^{1} (2[w+1])^{-1}\exp\{2[w+1]\}\underbrace{\mathbb{E}^U[\exp(-\log(U)/2[w+1])]}_{X=-\log(U)/2[w+1],\ U\sim\cal{U}(0,1)}2\text{d}w\\ &=\int_{0}^{1} (2[w+1])^{-1}\exp\{2[w+1]\}\,\mathbb{E}^U[U^{-1/2[w+1]}]\,2\text{d}w\end{align*}

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