Recent content by ariberth

Haha yes, thank you :)

Ah ok i think i found it allready :) $$\int_a^b \int_c^{f(x)} g(r)dr dx = \int_a^b |_c^{f(x)} G(r) = \int_a^b G(f(x)) - G(c) dx = \int_a^b G(f(x))dx - G(c) $$

Hello everybody! Are there methods to solve integrals of the following form? $$\int_a^b \int _0 ^{f(x)} g(r) dr dx$$ ariberth

I don't believe it. So the variance allways aproaches 0, no matter what choice of g i take?

$$\sum_{i=0}^{n-1} \mathcal{N}(f(t_i),g(t_i)) \Delta t = \sum_{i=0}^{3} \mathcal{N}(i\frac{1}{4},1) \frac{1}{2} =\frac{1}{4}(\mathcal{N}(0,1) + \mathcal{N}(\frac{1}{4},1) + \mathcal{N}(\frac{2}{4},1) +\mathcal{N}(\frac{3}{4},1)) = \frac{1}{4} \mathcal{N}(\frac{6}{4},4) =...

So that means i have to solve the following two integralls: $$\lim\limits_{i \to \infty}\sum_{i=0}^{n-1}\Delta tf(t_i)$$ and $$ \lim\limits_{i \to \infty}\sum_{i=0}^{n-1}(\Delta t) ^2g(t_i)$$ The first one is easy since: $$\lim\limits_{i \to \infty}\sum_{i=0}^{n-1}f(t_i) \Delta t= \int_a^b...

Thanks a lott for the tip with the rieman sums. Following your hint i discovered that i can use the fact that the normal is closed under linear transformation. So for the first example where $a=0, b=1, f(t)=t, g(t)=1$, and $n=2$ this leads to: $$\sum_{i=0}^{n-1} \mathcal{N}(f(t_i),g(t_i)) \Delta...

Hallo math helpers , i am trying to understand how one could solve the following integrall: $$\int_a^b \mathcal{N}(f(x_1,...,x_n,t),g(x_1,...,x_n,t)) dt$$, where $$\mathcal{N}$$ is the normal distribution, and $$f(x_1,...,x_n,t): \mathbb{R}^{n+1} \rightarrow \mathbb{R}$$, $$g(x_1,...,x_n,t)...