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We measure voltage across a resistor and find that the measured samples are distributed normally with mean $E[X]$ and variance $\sigma_X$, we measure current across a resistor and find that the measured samples are distributed normally with mean $E[Y]$ and variance $\sigma_Y$. It is given that both measurements are independent of each other. Random variables X and Y are independent.

What is the mean and variance of Z which represents the resistance? Since $V = IR$,

$$Z = \frac{X}{Y}$$

This question is inspired from the GATE 2019 Instrumentation paper which had the following question,

GATE IN 2019

So I would like to specifically find the $Var(Z)$ if $E[X] = 1$, $E[Y] = 10^{-3}$, $Var(X) = (0.12)^2$ and $Var(Y) = (0.05 \times 10^{-3})^2$

My attempts

From my googling, I find that

There is no formula expressing $E[\frac{1}{Y}]$ or $var(\frac{1}{Y})$ in terms of $E[Y]$ and $var(Y)$

So my initial attempts as shown below failed.

$$var(XY)=E(X^2Y^2)−E(XY)^2=var(X)var(Y)+var(X)E(Y)^2+var(Y)E(X)^2$$

The pdf is not given to us, so I can't manually integrate and find the expectation.

Since this question appeared in a prestigious national exam, I am confident a solution exists. Further this question was only worth 1 mark so I am expecting an elegant solution to exist.


Can we think of standard deviation as percent error? and simply add it?

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  • $\begingroup$ What level is the exam? They might just want you to apply basic error propagation... $\endgroup$ – Thomas Feb 8 at 13:47
  • $\begingroup$ It is meant for recently graduated students. GATE. I think you are right, it simply needs adding of the error I think. I can't believe I did not think of that. :( $\endgroup$ – Aditya Feb 8 at 13:50
  • $\begingroup$ Would and approximation be also good enough? $\endgroup$ – Math-fun Feb 8 at 13:57
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    $\begingroup$ For the record, no normal random variable $Y$ with nonzero variance (that is, not Dirac) is integrable. Thus, $E(1/Y)$, $E(1/Z)$, $\mathrm{var}(Y)$ and $\mathrm{var}(Z)$ are all undefined. GATE 2019 asking this is a display of incompetence. $\endgroup$ – Did Feb 8 at 14:19
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    $\begingroup$ Mmm... For example here en.m.wikipedia.org/wiki/Propagation_of_uncertainty look at the two variable example . It is essentially what J.G. proposes as the delta method, but such formulas are usually presented in lab courses via intuitive approaches $\endgroup$ – Thomas Feb 9 at 15:17
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What you're expected to do here is use the delta method, affecting the approximation $\operatorname{Var}f(X_i)\approx\sum_i\left(\frac{\partial f}{\partial X_i}\right)^2\operatorname{Var}X_i$. In this case, $\sigma_Z^2\approx\frac{\sigma_X^2}{Y^2}+\frac{X^2\sigma_Y^2}{Y^4}$.

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  • $\begingroup$ For the record you mean $E[X]^2$ when you say $X^2$ right? So that comes to $\sigma_Z = \sqrt{\frac{0.12^2}{10^-6} + \frac{1^2(0.05 \times 10^{-3})^2}{10^{-12}}} = 130$ $\endgroup$ – Aditya Feb 8 at 15:00

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