Calculating probability of displacement using two CDFs My knowledge of stats is fairly basic, so you please bear with me!
I'm trying to calculate the CDF for the vertical displacement of a (light, small) object floating in a wave tank. I have experimental data which suggests that the displacement can be approximated to a normal distribution, however I am also required to provide an analytical solution, as the experimental data is not reliable enough. 
My approach so far has been to use the Rayleigh Distribution to calculate the probability of any given wave being greater than an amplitude (h). 
$$P(H>h) = e^{-2\left(\frac{h^2}{h_s^2}\right)}$$
where $$ h_s = 2\sigma $$
Separately, I have approximated each individual wave to a sin wave of constant frequency. I have calculated the CDF of a sin wave (of amplitude H) to be:
$$P(X>x) = \frac{1}{2}-\frac{sin^{-1}(\frac{x}{H})}{\pi}$$
(where x is the vertical displacement of the object)
The part I do not understand, is how I combine these two to get the probability of exceeding a certain displacement for any wave. Instinctively, I suspect I am supposed to multiply the two and integrate for h between x and inf, but so far I haven't found an answer which is anywhere near what I am expecting, so I may be way off!
Thanks in advance!
 A: Try to define the probability that you want to obtain in mathematical terms. If you do so the following becomes just calculus.
The probability you want to obtain can be seen as an expectation of $I\left[x > x_0\right]$ over joint distribution of $x$ and $h$ where $I$ is an indicator function that is equal to $1$ when $x > x_0$ and to $0$ otherwise:
$$P(x > x_0) = \mathop{\mathbb E} _{x, h} I\left[x > x_0\right] = 
\int dP(x, h)\ I\left[x > x_0\right] = 
\iint dx\ dh\ p(x, h)\ \theta(x - x_0).$$
In this expression $p(x,h)$ is a probability density of joint distribution of $x$ and $h$, and $\theta(x)$ is a Heaviside Step function.
Your probability $P(X > x)$ is actually a conditional probability for given value of $h$ and its derivative is a probability density of conditional probability of $x$ for given $h$ $p(x \mid h).$ 
So using definition of conditional probability you can get that
$p(x, h)\ dx\ dh = p(x \mid h)\ dx \cdot p(h)\ dh.$
So now you can obtain the probability density functions
$$p(x \mid h) = \frac{d}{dx} P(X \leq x \mid h) = \frac{d}{dx}\left(1 - P(X > x \mid h) \right) = -\frac{\partial}{\partial x}P(X > x \mid h),$$
$$p(h) = \frac{d}{dh} P(H \leq h) = -\frac{\partial}{\partial h} P(H > h)$$
and substitute them to the integral:
$$P(x > x_0) = \iint dx\ dh\ p(x\mid h)\ p(h)\ \theta(x - x_0)=
\int\limits_0^{\infty} dh\ p(h) \int\limits_{x_0}^{\infty} dx\ p(x\mid h) = \\-\int\limits_0^{\infty} dh\ P(X > x_0 \mid h) \frac{\partial}{\partial h} P(H > h).$$

Actually in this particular case there is a more simple way to use conditional probabilities directly:
$$P(X > x) = \mathop{\mathbb E}_h P(X > x \mid h) = \int P(X > x\mid h) \ dP(h),$$
but I find the joint distribution/indicator function approach to be more powerful because it allows to infer probability distribution of more complicated relations (e.g. if probability of wave energy $x^2$ is larger than given function of $h$) or introduce new probabilistic variables to the model (e.g. instead of single value of $\sigma$ use random variable sigma from distribution obtained in result of bayesian estimation of $\sigma$ from experimental data) using a single generic framework.
