Inferences about $\mu$ based on the sum of two dependent normal RVs

Question

Given:

$X \sim \mathcal{N}(\mu, \sigma^2)$
$Y|X=x \sim \mathcal{N}(0, (\theta x)^2)$
$Z = X + Y$

I want to be able to make hypothesis tests or confidence intervals for $\mu$ using $Z$ and known $\theta$ and $\sigma^2$. For example, I draw $z = 950$, and I have $\sigma^2 = 10000$ and $\theta = 0.10$. How would I estimate a confidence interval for $\mu$ with this information?

Edit:

I'm not really sure how to go about this analytically, but I've been able to create confidence intervals numerically. Here's an empirical plot of the likelihood of $\mu$ given the observed $z$ and parameters in the above example.

Code for the above plot:

import scipy.stats as stats
import numpy as np
import pandas as pd
x_var   = 10000
theta   = 0.10
samples = int(4e7)
df1       = pd.DataFrame(index=range(samples))
df1['mu'] = stats.uniform.rvs(size=samples)1500+500
df1['x']  = stats.norm.rvs(loc=df1['mu'] , scale=np.sqrt(x_var)         , size=samples)
df1['y']  = stats.norm.rvs(loc=0         , scale=np.abs(thetadf1['x']) , size=samples)
df1['z']  = df1['x'] + df1['y']
z_observed = 950
df1.drop(df1[np.round(df1['z']) != z_observed].index, inplace = True)
hist1 = (pd.DataFrame(np.histogram(df1['mu'],bins=99,density=True))
            .transpose().rename(columns={0: 'L_mu', 1: 'mu'}))
hist1.plot(x='mu',y='L_mu')

Answer 1

Writing the density of $X$ as $f_X(x)$ and the conditional density of $Y| X=x$ as $f_{Y|x}(y|x)$ we find the joint density as $$ \frac{{\mathrm e}^{\frac{-\left(-x +\mu \right)^{2} \theta^{2} x^{2}-y^{2} \sigma^{2}}{2 \sigma^{2} \theta^{2} x^{2}}}}{2 \sigma \pi \theta {| x |}} $$ which I cannot recognize as a known distribution. We can try to find the density of $Z=X+Y$ by transformation to $(Z=X+Y, Y)$ (which have a jacobian of 1) and then integrating out $y$, I find the joint density of $(Z,Y)$ to be $$ \frac{{\mathrm e}^{\frac{-\left(-z +y +\mu \right)^{2} \left(-z +y \right)^{2} \theta^{2}-y^{2} \sigma^{2}}{2 \sigma^{2} \left(-z +y \right)^{2} \theta^{2}}}}{2 \sigma \pi \theta {| -z +y |}} $$ but trying to integrate out $y$ does not give a closed form: $$ \int_{-\infty}^{\infty}\frac{{\mathrm e}^{\frac{-\left(-z +y +\mu \right)^{2} \left(-z +y \right)^{2} \theta^{2}-y^{2} \sigma^{2}}{2 \sigma^{2} \left(-z +y \right)^{2} \theta^{2}}}}{2 \sigma \pi \theta {| -z +y |}}d y $$ but you could use this expression for finding the likelihood via numerical integration. See Parameter Estimation for intractable Likelihoods / Alternatives to approximate Bayesian computation for other ideas for intractable likelihoods.

Another idea is to calculate the moment generating function, which do have a closed form. It can be found relatively easily via the double expectation theorem (again I jump the details, I did those with maple): $$ \DeclareMathOperator{\E}{\mathbb{E}} M_Z(t)= \E e^{tZ} = \frac{\exp\left\{ -\frac{t(\mu^2 t \theta^2+\sigma^2 t + 2\mu)}{2(\sigma^2 t^2 \theta^2 -1)} \right\}}{\sqrt{1-\sigma^2 t^2 \theta^2}} $$ which is valid as long as the argument of the square root is positive, which is for $t^2 < \frac1{\sigma^2 \theta^2}$. Then we can use the saddlepoint approximation, see How does saddlepoint approximation work?, as an approximate likelihood function.

Out of time now, will add details later.