Hector Petersen

2022-06-25

'Alice is an astronomer performing many measurements which she averages to estimate the true distance $d$ of a star. Assume the measurements are independent, each with mean $d$. Alice is slowly getting more and more tired. Hence her $n$th measurement has standard deviation ${n}^{1/3}$. Let $n$ be the number of times Alice repeats measuring this distance. As $n$ goes to infinity, can she expect her average measurement to converge to the actual distance $d$?''

I'm pretty stuck for this question, because it seems like I would want to invoke the Strong Law of Large Numbers, but the conditions do not hold since each measurement is not identically distributed. Do I have to manipulate these variables somehow to be able to use the SLLN, or am I missing something else entirely. Another idea I had, was that maybe it was possible that I would only need to use some form of the Weak Law of Large Numbers, which would be more easy to use, but I'm not sure if the WLLN would actually be a strong enough conclusion anyway for this quesiton (is this right?). Any help would be massively appreciated

I'm pretty stuck for this question, because it seems like I would want to invoke the Strong Law of Large Numbers, but the conditions do not hold since each measurement is not identically distributed. Do I have to manipulate these variables somehow to be able to use the SLLN, or am I missing something else entirely. Another idea I had, was that maybe it was possible that I would only need to use some form of the Weak Law of Large Numbers, which would be more easy to use, but I'm not sure if the WLLN would actually be a strong enough conclusion anyway for this quesiton (is this right?). Any help would be massively appreciated

assumintdz

Beginner2022-06-26Added 22 answers

Let the $k$th measurement be ${X}_{k}$. To see that this converges to $d$, we can see that

$\text{Var}\left(\frac{1}{n}\sum _{k=1}^{n}{X}_{k}\right)\to 0$

as $n\to \mathrm{\infty}$. We have

$\begin{array}{rl}\text{Var}\left(\frac{1}{n}\sum _{k=1}^{n}{X}_{k}\right)& =\frac{1}{{n}^{2}}\text{Var}\left(\sum _{k=1}^{n}{X}_{k}\right)\\ \text{(by independence)}& & =\frac{1}{{n}^{2}}\sum _{k=1}^{n}\text{Var}({X}_{k})& =\frac{1}{{n}^{2}}\sum _{k=1}^{n}{k}^{\frac{2}{3}}\\ & \le \frac{1}{{n}^{2}}\sum _{k=1}^{n}{n}^{\frac{2}{3}}\\ & =\frac{1}{{n}^{2}}n\cdot {n}^{\frac{2}{3}}\\ & ={n}^{-\frac{1}{3}}\to 0.\end{array}$

$\text{Var}\left(\frac{1}{n}\sum _{k=1}^{n}{X}_{k}\right)\to 0$

as $n\to \mathrm{\infty}$. We have

$\begin{array}{rl}\text{Var}\left(\frac{1}{n}\sum _{k=1}^{n}{X}_{k}\right)& =\frac{1}{{n}^{2}}\text{Var}\left(\sum _{k=1}^{n}{X}_{k}\right)\\ \text{(by independence)}& & =\frac{1}{{n}^{2}}\sum _{k=1}^{n}\text{Var}({X}_{k})& =\frac{1}{{n}^{2}}\sum _{k=1}^{n}{k}^{\frac{2}{3}}\\ & \le \frac{1}{{n}^{2}}\sum _{k=1}^{n}{n}^{\frac{2}{3}}\\ & =\frac{1}{{n}^{2}}n\cdot {n}^{\frac{2}{3}}\\ & ={n}^{-\frac{1}{3}}\to 0.\end{array}$

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