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}]]]
$$\text{NormalDistribution}\left[0,\frac{1}{4} \sqrt{\frac{1}{10} (120+8 \sin (20)+\sin (40))}\right] $$
Sin[4x]
and standard deviationCos[10x]^2
and $X$ has a uniform distribution on $(-1,1)$ ? $\endgroup$dist(x)
represent a normal pdf? $\endgroup$