# Describing the measurement of a random variable as another random variable

Background

Suppose we have a box of resistors. The manufacturer rates these resistors at 100 ohms, but they have some variability. Let $$x$$ be the true resistance of a resistor chosen from the box at random. Suppose

$$x \sim N(\mu,\sigma^2)$$

We have an ohmmeter that can measure the resistance of the selected resistor. Denote the measurement of the resistor as $$y=x+\epsilon$$, where $$\epsilon \sim N(0,\delta^2)$$

In the textbook I'm studying, they have the expression $$Q = E\left[(x-\hat{x})^2|y\right]$$ where $$\hat{x}$$ is an estimate for the true resistance $$x$$.

The goal is to take the derivative of $$Q$$ with respect to $$\hat{x}$$ and set to zero to find the optimum estimate $$\hat{x}$$. I was working out the details for this derivative/expectation and found that the "Law of the Unconscious Statistician" might be useful. I tried applying this law and ran into the conditional distribution $$f_{x|y}(x|y)=f(x,y)/f(y)$$. Then I realized I wasn't sure what $$f(y)$$ was exactly.

Question

My question is regarding how to describe the distribution of the measurement $$y$$. If we think of all possible measurements of all possible resistors from the box, I think we would write

$$y \sim N(\mu,\sigma^2+\delta^2)$$

However, if we are thinking of the possible measurements for a single, randomly selected resistor with true resistance $$x$$, then we would write

$$y\sim N(x,\delta^2)$$

Which distribution for $$y$$ do I care about when I'm trying to find the conditional distribution $$f(x|y)$$? And is there better notation to use so that I can avoid this confusion?

• As you've stated, the first distribution of $y$ is when we don't know the true resistance $x$, and the second distribution of $y$ is when we're given $x$, which can be written as $y|x \sim N(x, \delta^2)$ (I need someone to verify my using of this notation though). By the way, the background problem you're considering doesn't seem to ask for the distribution of $y$ or $y|x$, because we're given $y$. It'll be straightforward to think about the distribution of $x$ given $y$, or $x|y$. Jun 15, 2022 at 4:23
• Apply Bayes' theorem:$$f(x,y)=f(x)f(y|x)=f(y)f(x|y)$$ Jun 15, 2022 at 8:15