MHB Integral over gaussian pdf where parameters depend on integrand

[ariberth]

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 $$f(x_1,...,x_n,t): \mathbb{R}^{n+1} \rightarrow \mathbb{R}$$, $$g(x_1,...,x_n,t): \mathbb{R}^{n+1} \rightarrow
\mathbb{R}$$. So the mean and variance changes with t . I read that $$\mathcal{N}(\mu_1,\sigma_1) + \mathcal{N}(\mu_2,\sigma_2) = \mathcal{N}(\mu_1 + \mu_2,\sigma_1+\sigma_2)$$ Does that mean that i just need to integrate f and g respectively?

 

[I like Serena]

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 $$f(x_1,...,x_n,t): \mathbb{R}^{n+1} \rightarrow \mathbb{R}$$, $$g(x_1,...,x_n,t): \mathbb{R}^{n+1} \rightarrow
\mathbb{R}$$. So the mean and variance changes with t .

Hi ariberth! Welcome to MHB! (Smile)

Since $x_1,...,x_n$ are not referenced anywhere, we can assume them to be constant and reduce the problem to:
$$\int_a^b \mathcal{N}(f(t),g(t)) dt$$

I read that $$\mathcal{N}(\mu_1,\sigma_1) + \mathcal{N}(\mu_2,\sigma_2) = \mathcal{N}(\mu_1 + \mu_2,\sigma_1+\sigma_2)$$ Does that mean that i just need to integrate f and g respectively?

Not quite. It's a little more complex.
It should be:
$$\mathcal{N}(\mu_1,\sigma_1) + \mathcal{N}(\mu_2,\sigma_2) = \mathcal{N}(\mu_1 + \mu_2,\sqrt{\sigma_1^2+\sigma_2^2})$$
or with an alternative and easier notation:
$$\mathcal{N}(\mu_1,\sigma_1^2) + \mathcal{N}(\mu_2,\sigma_2^2) = \mathcal{N}(\mu_1 + \mu_2,\sigma_1^2+\sigma_2^2)$$We can write the integral as the limit of, say, a Left Riemann Sum (see the definition of a Riemann integral):
$$\int_a^b \mathcal{N}(f(t),g(t)) dt = \lim_{n \to \infty} \sum_{i=0}^{n-1} \mathcal{N}(f(t_i),g(t_i)) \Delta t
$$
where $\Delta t = \frac{b-a}n$ and $t_i = a + i\Delta t$.Let's pick an example.

Suppose we pick $a=0, b=1, f(t)=t, g(t)=1$, and $n=2$.
What will be the Riemann sum:
$$\sum_{i=0}^{n-1} \mathcal{N}(f(t_i),g(t_i)) \Delta t$$
? (Wondering)

And what if we pick $n=4$?

 

[ariberth]

Let's pick an example.

Suppose we pick $a=0, b=1, f(t)=t, g(t)=1$, and $n=2$.
What will be the Riemann sum:
$$\sum_{i=0}^{n-1} \mathcal{N}(f(t_i),g(t_i)) \Delta t$$
? (Wondering)

And what if we pick $n=4$?

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:
$$\sum_{i=0}^{n-1} \mathcal{N}(f(t_i),g(t_i)) \Delta t = \sum_{i=0}^{1} \mathcal{N}(i\frac{1}{2},1) \frac{1}{2} = (\mathcal{N}(0,1) + \mathcal{N}(\frac{1}{2},1)) \frac{1}{2} = \mathcal{N}(\frac{1}{2},2) \frac{1}{2} = \mathcal{N}(\frac{1}{4},\frac{1}{2})$$
Using that, in the limit this would lead to: $$\lim_{x \to \infty} \sum_{i=0}^{n-1} \mathcal{N}(f(t_i),g(t_i)) \Delta t = \lim_{x \to \infty} \mathcal{N}( \sum_{i=0}^{n-1}\Delta tf(t_i), \sum_{i=0}^{n-1}(\Delta t) ^2g(t_i)) $$ Is that correct?

 

[I like Serena]

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:
$$\sum_{i=0}^{n-1} \mathcal{N}(f(t_i),g(t_i)) \Delta t = \sum_{i=0}^{1} \mathcal{N}(i\frac{1}{2},1) \frac{1}{2} = (\mathcal{N}(0,1) + \mathcal{N}(\frac{1}{2},1)) \frac{1}{2} = \mathcal{N}(\frac{1}{2},2) \frac{1}{2} = \mathcal{N}(\frac{1}{4},\frac{1}{2})$$
Using that, in the limit this would lead to: $$\lim_{x \to \infty} \sum_{i=0}^{n-1} \mathcal{N}(f(t_i),g(t_i)) \Delta t = \lim_{x \to \infty} \mathcal{N}( \sum_{i=0}^{n-1}\Delta tf(t_i), \sum_{i=0}^{n-1}(\Delta t) ^2g(t_i)) $$ Is that correct?

Yep - assuming that g(t) is the variance instead of the standard deviation. (Nod)

 

[ariberth]

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: $$\lim\limits_{i \to \infty}\sum_{i=0}^{n-1}f(t_i) \Delta t= \int_a^b f(t) dt$$
and i just have to find the anti-derivative of f.

Is there a way to do the same thing with the second:
$$ \lim\limits_{i \to \infty}\sum_{i=0}^{n-1}g(t_i) (\Delta t) ^2 = ??$$

 

[I like Serena]

Good!

Let's do another example.
Or rather, the same example with n=4.
What's the pattern?

 

[ariberth]

$$\sum_{i=0}^{n-1} \mathcal{N}(f(t_i),g(t_i)) \Delta t = \sum_{i=0}^{3} \mathcal{N}(i\frac{1}{4},1) \frac{1}{2} =\frac{1}{4}(\mathcal{N}(0,1) + \mathcal{N}(\frac{1}{4},1) + \mathcal{N}(\frac{2}{4},1) +\mathcal{N}(\frac{3}{4},1)) = \frac{1}{4} \mathcal{N}(\frac{6}{4},4) = \mathcal{N}(\frac{6}{16},\frac{4}{16})$$ The only pattern i see is that the scalar from $$\Delta t$$ is the inverse of the variance but that depends on how g is chosen. So i don't really know what the pattern is...

 

[I like Serena]

$$\sum_{i=0}^{n-1} \mathcal{N}(f(t_i),g(t_i)) \Delta t = \sum_{i=0}^{3} \mathcal{N}(i\frac{1}{4},1) \frac{1}{2} =\frac{1}{4}(\mathcal{N}(0,1) + \mathcal{N}(\frac{1}{4},1) + \mathcal{N}(\frac{2}{4},1) +\mathcal{N}(\frac{3}{4},1)) = \frac{1}{4} \mathcal{N}(\frac{6}{4},4) = \mathcal{N}(\frac{6}{16},\frac{4}{16})$$ The only pattern i see is that the scalar from $$\Delta t$$ is the inverse of the variance but that depends on how g is chosen. So i don't really know what the pattern is...

The pattern is that $\mu$ approaches $\frac 1 2$ as expected, since $\int f(t)dt = \int_0^1 t\,dt = \frac 12$.
And we see that $\sigma^2$ becomes smaller and smaller, approaching 0.

Indeed, $\sum g(t_i) \Delta t$ approaches $\int g(t)dt$.
So multiplying it with another $\Delta t$ makes it approach 0.

This is equivalent to the fact that when you take averages long enough (ad infinitum), finally you will be left with the expected mean and negligible variance.

 

[ariberth]

The pattern is that $\mu$ approaches $\frac 1 2$ as expected, since $\int f(t)dt = \int_0^1 t\,dt = \frac 12$.
And we see that $\sigma^2$ becomes smaller and smaller, approaching 0.

Indeed, $\sum g(t_i) \Delta t$ approaches $\int g(t)dt$.
So multiplying it with another $\Delta t$ makes it approach 0.

This is equivalent to the fact that when you take averages long enough (ad infinitum), finally you will be left with the expected mean and negligible variance.

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

 

[I like Serena]

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

Yup.

Well... maybe if you have a $g$ that approaches infinity...
... or take an interval that is infinitely large...