Normal Distribution with Normally Distributed Mean

Question

I am looking to determine the distribution of $X$, given:

$Y \sim N(29.5, 6.49)$

$X \sim N(y, 0.16)$

Where $y$ is sampled from $Y$.

edit: I am working on a project attempting to determine a confidence regarding the achieved soil strength. I have an assumed distribution of the strength, and know the variance of the test procedure. To update the in-situ soil strength, I am looking to determine a distribution for the test strength: ie. test strength is comparatively a "sample" of the normally distributed in-situ strength.

Answer 1

The simplest way (IMO) is to use the characteristic function approach, along with conditional expectation. $$\mathbb{E}[e^{itX}]=\mathbb{E}[\mathbb{E}[e^{itX}|Y]]$$ and use the known expression of the cf for Gaussians. This yields the answer with one line of derivation.

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In more detail: (put your mouse over the hidden text to reveal it)

Let $\mu = 29.5$, $\sigma_Y=6.49$, $\sigma_X=0.16$. for every $t\in\mathbb{R}$, $$\mathbb{E}[e^{itX}]=\mathbb{E}[\mathbb{E}[e^{itX}|Y]]=\mathbb{E}[e^{itY-\frac{1}{2}\sigma_X^2t^2}]=e^{-\frac{1}{2}\sigma_X^2t^2}\mathbb{E}[e^{itY}]=e^{-\frac{1}{2}\sigma_X^2t^2}e^{it\mu-\frac{1}{2}\sigma_Y^2t^2}=\boxed{e^{it\mu-\frac{1}{2}(\sigma_Y^2+\sigma_X^2)t^2}} $$so $X$ is distributed as a Normal with mean $\mu$ and variance $\sigma^2 = \sigma_Y^2+\sigma_X^2$.

Answer 2

The density function for $X$ is $f_X(x)=\frac{1}{2\pi \sigma_x\sigma_y}\int_{-\infty}^{\infty} e^{-\frac{(x-y)^2}{2\sigma_x^2}}e^{-\frac{(y-\mu_y)^2}{2\sigma_y^2}}dy$.

Here $\mu_y=29.5$, $\sigma_y=6.49$, and $\sigma_x=0.16$