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.


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?

  • $\begingroup$ 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 $x|y$. $\endgroup$ Jun 15, 2022 at 4:23
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    $\begingroup$ Apply Bayes' theorem:$$f(x,y)=f(x)f(y|x)=f(y)f(x|y)$$ $\endgroup$
    – Xi'an
    Jun 15, 2022 at 8:15


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