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If $y_{1}, y_{2}, . . . , y_{N}$form a sample of independent standard-normally distributed random variables and $\bar{y}$ is the sample average.

Is it correct to say that $$\bar{y}^2 \overset{p}{\rightarrow} E[\bar{y}^2] =0$$

user407052
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    Because the expectation of a square is positive (unless the square is almost surely zero), your guess fails at the final equality. – whuber Feb 20 '24 at 19:40

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Two points:

  • $E[\bar y^2]=E[\bar y]^2+\mathrm{var}[\bar y]$. The first term of this is the zero that you have, the second term is non-zero, as @whuber suggested
  • This is exactly the sort of question where simulation is helpful (Math is hard; let's use science). It's really easy to use simulation to check your guesses/hypotheses -- you do want to guess first, otherwise you won't know whether to be surprised
> r<-replicate(1000, mean(rnorm(10000)^2))
> summary(r)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.9570  0.9915  1.0003  1.0006  1.0101  1.0389
Thomas Lumley
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