Return to Answer

The fundamental validation of the Monte Carlo method is the Law of Large Numbers: when $𝔼[X]$ does exist, meaning when $𝔼[|X|]$ is finite, then the empirical average$$\bar{X}_T=\frac{1}{T}\sum_{t=1}^T X_t$$converges almost surely to $𝔼[X]$.

When the expectation $𝔼[X]$ does not exist, meaning$$\int |x|f(x)\text{d}x=+\infty$$there is no guarantee for the empirical average to converge, see e.g. the case of the iid standard Cauchy sequence, when $\bar{X}_T$ remains a standard Cauchy for all $T$'s. But there also exist cases when it converges, as discussed in another XV question:

Counter-examples provided in Wikipedia are

1. $X=\sin(Z)\exp\{Z\}/Z$ when $Z\sim\mathcal{E}xp(1)$, with $\mu=\pi/2$
2. $X=2^Z(-1)^Z/z$ when $Z\sim\mathcal{G}(1/2)$, with $\mu=-\log(2)$
3. $X\sim F(x)$ with $$F(x)=\mathbb{I}_{x\ge e}-\frac{e\mathbb{I}_{x\ge e}}{2x\log(x)}-\frac{e\mathbb{I}_{x\le-e}}{2x\log(-x)}+\frac{\mathbb{I}_{-e\le x\le e}}{2}$$with $\mu=0$

[in the sense that the rv's have no expectation but there exists a limit $\mu$$-$in probability if not a.s.$-$for the sample average $\bar{X}_n$].

See also this quite informative answer on XV.

Now, if you take an easily divergent example such as the Pareto $\cal{P}(1/2,4)$ distribution $$f(x)=x^{-3/2}\mathbb{I}_{x>4}$$ simulating this distribution is equivalent to turn a uniform $U$ into $4/U^2$. The mean of the Pareto $\cal{P}(1/2,4)$ distribution is infinite: $$\int_4^\infty x^{-1/2}\text{d}x=+\infty$$But a Monte Carlo experiment does not exhibit a lack of convergence or convergence to $+\infty$, simply that the average can take arbitrarily large values after any number of iterations, as demonstrated on this experiment with 10⁶ simulations, repeated 10² times:

The fundamental validation of the Monte Carlo method is the Law of Large Numbers: when $𝔼[X]$ does exist, meaning when $𝔼[|X|]$ is finite, then the empirical average$$\bar{X}_T=\frac{1}{T}\sum_{t=1}^T X_t$$converges almost surely to $𝔼[X]$.

When the expectation $𝔼[X]$ does not exist, meaning$$\int |x|f(x)\text{d}x=+\infty$$there is no guarantee for the empirical average to converge, see e.g. the case of the iid standard Cauchy sequence, when $\bar{X}_T$ remains a standard Cauchy for all $T$'s. But there also exist cases when it converges, as discussed in [another XV question][1]:

Counter-examples provided in [Wikipedia][2] are

1. $X=\sin(Z)\exp\{Z\}/Z$ when $Z\sim\mathcal{E}xp(1)$, with $\mu=\pi/2$
2. $X=2^Z(-1)^Z/z$ when $Z\sim\mathcal{G}(1/2)$, with $\mu=-\log(2)$
3. $X\sim F(x)$ with $$F(x)=\mathbb{I}_{x\ge e}-\frac{e\mathbb{I}_{x\ge e}}{2x\log(x)}-\frac{e\mathbb{I}_{x\le-e}}{2x\log(-x)}+\frac{\mathbb{I}_{-e\le x\le e}}{2}$$with $\mu=0$

[in the sense that the rv's have no expectation but there exists a limit $\mu$$-$in probability if not a.s.$-$for the sample average $\bar{X}_n$].

See also [this quite informative answer][3] on XV.

Now, if you take an easily divergent example such as the Pareto $\cal{P}(1/2,4)$ distribution $$f(x)=x^{-3/2}\mathbb{I}_{x>4}$$ simulating this distribution is equivalent to turn a uniform $U$ into $4/U^2$. The mean of the Pareto $\cal{P}(1/2,4)$ distribution is infinite: $$\int_4^\infty x^{-1/2}\text{d}x=+\infty$$But a Monte Carlo experiment does not exhibit a lack of convergence or convergence to $+\infty$, simply that the average can take arbitrarily large values after any number of iterations, as demonstrated on this experiment with 10⁶ simulations, repeated 10² times:

[![enter image description here][4]][4]

Note also that, in the case of positive rv's like the Pareto $\cal{P}(1/2,4)$ distribution above, $\bar{X}_T$ does not have to converge to infinity if $\mathbb{E}[X]=\infty$. Indeed by Chebychev's inequality, \begin{align*} \mathbb{P}(\bar{X}_T\ge a) &\le \frac{1}{a^\epsilon}\int_{x>a} x^\epsilon \text{d}\mathbb{P}^{\bar{X}_T}(x) \end{align*} which may be finite for $\epsilon>0$ small enough, in which case the rhs probability goes to zero as $a$ goes to infinity. [1]: https://stats.stackexchange.com/a/328039/7224 [2]: https://en.wikipedia.org/wiki/Law_of_large_numbers#Differences_between_the_weak_law_and_the_strong_law [3]: https://stats.stackexchange.com/a/29891/7224 [4]: https://i.sstatic.net/i3z8S.jpg

The fundamental validation of the Monte Carlo method is the Law of Large Numbers: when $𝔼[X]$ does exist, meaning when $𝔼[|X|]$ is finite, then the empirical average$$\frac{1}{T}\sum_{t=1}^T X_t$$converges almost surely to $𝔼[X]$.

When the expectation $𝔼[X]$ does not exist, meaning$$\int |x|f(x)\text{d}x=+\infty$$there is no guarantee for the empirical average to converge, see e.g. the case of the iid standard Cauchy sequence, when $\bar{X}_n$ remains a standard Cauchy for all $n$'s, although there exist cases when it converges. As discussed in another XV question:

Counter-examples provided in Wikipedia are

1. $X=\sin(Z)\exp\{Z\}/Z$ when $Z\sim\mathcal{E}xp(1)$, with $\mu=\pi/2$
2. $X=2^Z(-1)^Z/z$ when $Z\sim\mathcal{G}(1/2)$, with $\mu=-\log(2)$
3. $X\sim F(x)$ with $$F(x)=\mathbb{I}_{x\ge e}-\frac{e\mathbb{I}_{x\ge e}}{2x\log(x)}-\frac{e\mathbb{I}_{x\le-e}}{2x\log(-x)}+\frac{\mathbb{I}_{-e\le x\le e}}{2}$$with $\mu=0$

[in the sense that the rv's have no expectation but there exists a limit $\mu$$-$in probability if not a.s.$-$for the sample average $\bar{X}_n$].

See also this quite informative answer on XV.

Now, if you take an easily divergent example such as the Pareto $\cal{P}(1/2,4)$ distribution $$f(x)=x^{-3/2}\mathbb{I}_{x>4}$$ simulating this distribution is equivalent to turn a uniform $U$ into $4/U^2$. The mean of the Pareto $\cal{P}(1/2,4)$ distribution is infinite: $$\int_4^\infty x^{-1/2}\text{d}x=+\infty$$But a Monte Carlo experiment does not exhibit a lack of convergence or convergence to $+\infty$, simply that the average can take arbitrarily large values after any number of iterations, as demonstrated on this experiment with 10⁶ simulations, repeated 10² times:

The fundamental validation of the Monte Carlo method is the Law of Large Numbers: when $𝔼[X]$ does exist, meaning when $𝔼[|X|]$ is finite, then the empirical average$$\bar{X}_T=\frac{1}{T}\sum_{t=1}^T X_t$$converges almost surely to $𝔼[X]$.

When the expectation $𝔼[X]$ does not exist, meaning$$\int |x|f(x)\text{d}x=+\infty$$there is no guarantee for the empirical average to converge, see e.g. the case of the iid standard Cauchy sequence, when $\bar{X}_T$ remains a standard Cauchy for all $T$'s. But there also exist cases when it converges, as discussed in another XV question:

Counter-examples provided in Wikipedia are

1. $X=\sin(Z)\exp\{Z\}/Z$ when $Z\sim\mathcal{E}xp(1)$, with $\mu=\pi/2$
2. $X=2^Z(-1)^Z/z$ when $Z\sim\mathcal{G}(1/2)$, with $\mu=-\log(2)$
3. $X\sim F(x)$ with $$F(x)=\mathbb{I}_{x\ge e}-\frac{e\mathbb{I}_{x\ge e}}{2x\log(x)}-\frac{e\mathbb{I}_{x\le-e}}{2x\log(-x)}+\frac{\mathbb{I}_{-e\le x\le e}}{2}$$with $\mu=0$

[in the sense that the rv's have no expectation but there exists a limit $\mu$$-$in probability if not a.s.$-$for the sample average $\bar{X}_n$].

See also this quite informative answer on XV.

Now, if you take an easily divergent example such as the Pareto $\cal{P}(1/2,4)$ distribution $$f(x)=x^{-3/2}\mathbb{I}_{x>4}$$ simulating this distribution is equivalent to turn a uniform $U$ into $4/U^2$. The mean of the Pareto $\cal{P}(1/2,4)$ distribution is infinite: $$\int_4^\infty x^{-1/2}\text{d}x=+\infty$$But a Monte Carlo experiment does not exhibit a lack of convergence or convergence to $+\infty$, simply that the average can take arbitrarily large values after any number of iterations, as demonstrated on this experiment with 10⁶ simulations, repeated 10² times:

The fundamental validation of the Monte Carlo method is the Law of Large Numbers: when $𝔼[X]$ does exist, meaning when $𝔼[|X|]$ is finite, then the empirical average$$\frac{1}{T}\sum_{t=1}^T X_t$$converges almost surely to $𝔼[X]$.

When the expectation $𝔼[X]$ does not exist, meaning$$\int |x|f(x)\text{d}x=+\infty$$there is no guarantee for the empirical average to converge, see e.g. the case of the Cauchy sequence, although there exist cases when it converges. As discussed in another XV question:

Counter-examples provided in Wikipedia are

1. $X=\sin(Z)\exp\{Z\}/Z$ when $Z\sim\mathcal{E}xp(1)$, with $\mu=\pi/2$
2. $X=2^Z(-1)^Z/z$ when $Z\sim\mathcal{G}(1/2)$, with $\mu=-\log(2)$
3. $X\sim F(x)$ with $$F(x)=\mathbb{I}_{x\ge e}-\frac{e\mathbb{I}_{x\ge e}}{2x\log(x)}-\frac{e\mathbb{I}_{x\le-e}}{2x\log(-x)}+\frac{\mathbb{I}_{-e\le x\le e}}{2}$$with $\mu=0$

[in the sense that the rv's have no expectation but there exists a limit $\mu$$-$in probability if not a.s.$-$for the sample average $\bar{X}_n$].

See also this quite informative answer on XV.

The fundamental validation of the Monte Carlo method is the Law of Large Numbers: when $𝔼[X]$ does exist, meaning when $𝔼[|X|]$ is finite, then the empirical average$$\frac{1}{T}\sum_{t=1}^T X_t$$converges almost surely to $𝔼[X]$.

When the expectation $𝔼[X]$ does not exist, meaning$$\int |x|f(x)\text{d}x=+\infty$$there is no guarantee for the empirical average to converge, see e.g. the case of the iid standard Cauchy sequence, when $\bar{X}_n$ remains a standard Cauchy for all $n$'s, although there exist cases when it converges. As discussed in another XV question:

Counter-examples provided in Wikipedia are

1. $X=\sin(Z)\exp\{Z\}/Z$ when $Z\sim\mathcal{E}xp(1)$, with $\mu=\pi/2$
2. $X=2^Z(-1)^Z/z$ when $Z\sim\mathcal{G}(1/2)$, with $\mu=-\log(2)$
3. $X\sim F(x)$ with $$F(x)=\mathbb{I}_{x\ge e}-\frac{e\mathbb{I}_{x\ge e}}{2x\log(x)}-\frac{e\mathbb{I}_{x\le-e}}{2x\log(-x)}+\frac{\mathbb{I}_{-e\le x\le e}}{2}$$with $\mu=0$

[in the sense that the rv's have no expectation but there exists a limit $\mu$$-$in probability if not a.s.$-$for the sample average $\bar{X}_n$].

See also this quite informative answer on XV.

Now, if you take an easily divergent example such as the Pareto $\cal{P}(1/2,4)$ distribution $$f(x)=x^{-3/2}\mathbb{I}_{x>4}$$ simulating this distribution is equivalent to turn a uniform $U$ into $4/U^2$. The mean of the Pareto $\cal{P}(1/2,4)$ distribution is infinite: $$\int_4^\infty x^{-1/2}\text{d}x=+\infty$$But a Monte Carlo experiment does not exhibit a lack of convergence or convergence to $+\infty$, simply that the average can take arbitrarily large values after any number of iterations, as demonstrated on this experiment with 10⁶ simulations, repeated 10² times: