Definite Integral of Probability Distributions

Question

Suppose at each value of $x$ I define a normal distribution

dist[x_] := NormalDistribution[Sin[4 x], Cos[10 x]^2]

Now I want to find the distribution

$$ \mathcal{D} = \int_{-1}^1 \text{dist}(x)\ dx$$

This is essentially what TransformedDistribution is meant for but it's usually for simple combinations of distributions (e.g TransformedDistribution[ x + y, {x \[Distributed] NormalDistribution[], y \[Distributed] NormalDistribution[]}]) not integrals over a continuous set of disributions.

Of course since $\text{dist}(x)$ is always normal I know that any sum of them is also normal which allows me to compute $\mathcal{D}$ exactly, but I wondering if theres some way to get Mathemtatica to output the exact result that uses all the power of TransformedDistribution.

Thanks

Exact solution should be

NormalDistribution[Integrate[Sin[4 x], {x, -1, 1}], 
 Sqrt[Integrate[Cos[10 x]^4, {x, -1, 1}]]]

$$\text{NormalDistribution}\left[0,\frac{1}{4} \sqrt{\frac{1}{10} (120+8 \sin (20)+\sin (40))}\right] $$

Answer 1

This can be done as follows. Let us consider the integral sums for $ \int_{-1}^1 \text{dist}(x)\, dx$, where the interval $[-1,1]$ is split into $n$ intervals of equal lengths and the values of $\text{dist}(x)$ are taken at its right ends:

ClearAll["Global`*"];
f[n_] := TransformedDistribution[Sum[2/n*dist[-1 + 2*j/n], 
{j, 1, n}], Table[dist[-1 + 2*j/n] \[Distributed] 
NormalDistribution[Sin[4 *(-1 + 2*j/n)], 
Cos[10 *(-1 + 2*j/n)]^2], {j, 1, n}]]

The result of

Table[f[n], {n, 1, 5}]

{NormalDistribution[2 Sin[4], 2 Cos[10]^2], NormalDistribution[Sin[4], Sqrt[1 + Cos[10]^4]], NormalDistribution[(2 Sin[4])/3, Sqrt[ 8/9 Cos[10/3]^4 + (4 Cos[10]^4)/9]], NormalDistribution[Sin[4]/2, Sqrt[1/4 + Cos[5]^4/2 + Cos[10]^4/4]], NormalDistribution[(2 Sin[4])/5, Sqrt[(8 Cos[2]^4)/25 + (8 Cos[6]^4)/25 + (4 Cos[10]^4)/25]]}

confirms that those sums are normal distributions. Now

FindSequenceFunction[Table[Mean[f[n]], {n, 1, 10}], n];

(2 Sin[4])/n

 DiscreteLimit[%,n->Infinity]

0

demonstrates that the mean of the limit distribution (which is normal according to certain theorems of probability theory) equals $0$. Unfortunately,

FindSequenceFunction[Table[Variance[f[n]], {n, 4, 14}], n]

returns the input. In view of it we have to somewhat help Mathematica. Making use of the known property of the variances of the sum of independent normal distributions, we obtain

DiscreteLimit[2/n*Sum[Variance[
NormalDistribution[Sin[4 *(-1 + 2*j/n)], 
 Cos[10 *(-1 + 2*j/n)]^2]], {j, 1, n}], n -> Infinity]

1/160 (120 + 8 Sin[20] + Sin[40])

Now

Simplify[Integrate[Cos[10 x]^4,{x,-1,1}]-%]

0

finishes the work.

Answer 2

If (from your comment) you know that a random variable $Y$ given the value of $X=x$ has a normal distribution with mean $\sin(4x)$ and standard deviation $\cos^2(10x)$ for $-1<x<1$:

$$Y|X=x \sim N(\sin(4x),\cos^2(10x))$$

Now suppose $X\sim U(-1,1)$ (i.e., $X$ has a uniform distribution between -1 and 1). The (unconditional) probability density function for $Y=y$ is

$$p(y)=\int_{-1}^1 \frac{e^{-\frac{(y-\sin (4 x))^2}{2\cos^4(10x)}}}{\sqrt{2 \pi }\cos^2(10x)} \times\frac{1}{2}dx$$

Other than the multiplier of $1/2$, this is what you show with

$$ \mathcal{D} = \int_{-1}^1 \text{dist}(x)\ dx$$

if $\text{dist}(x)$ is the conditional pdf of $Y|X=x$. (If $\text{dist}(x)$ is something else, then I don't know what that is without further explanation from you.)

There doesn't seem to be a nice closed-form for $p(y)$. But a numerical evaluation is the following:

Plot[NExpectation[PDF[NormalDistribution[Sin[4 x], Cos[10 x]^2], y],
  x \[Distributed] UniformDistribution[{-1, 1}]], {y, -4, 4},
 AxesLabel -> {"Y", "Probability density function"}]

A large random sample produces essentially the same pdf:

n = 1000000;
xx = RandomVariate[UniformDistribution[{-1, 1}], n];
parms = {Sin[4 #], Cos[10 #]^2} & /@ xx;
yy = RandomVariate[NormalDistribution[#[[1]], #[[2]]], 1][[1]] & /@ parms;
Histogram[yy, "FreedmanDiaconis", "PDF"]

In short, it seems that you might be asking about a weighted average of pdfs rather than a weighted average of random variables.