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$\bullet$ Prove: $\mathrm E(\bar x)= \mu$

Answer:

Let $x_1,x_2,x_3\ldots,x_n$ denote the sample observations. The sample mean is $$\bar x= \frac{(x_1+x_2+x_3+\ldots+x_n)}{n}= \frac{1}{n}\sum x_i$$ where $x_i$ is the $i$-th member of of the sample.

Note, in simple random sampling(with or without replacement), the sample members has the same probability distribution as in the variable $x$ in the population.

Therefore, $\mathrm E( x_i)= \mu$

And $$\mathrm E(\bar x)= \frac{1}{n}[\mathrm E(x_1)+ \mathrm E(x_2)+\ldots+\mathrm E(x_n)]= \mu.$$


What I'm not getting is the blocked part that the author wanted to highlight.

Can anyone tell me why actually $x_i$ has the same probability distribution as $x$ in the population especially even when the random sampling is done without replacement?

For $x_i$ and $x_j$ are not independent any-more when the sampling is done without replacement; so can then also the probability distribution of $x_i$ and $x_j$ remain the same as that of $x$ in the population?

How can $\mathrm E(x_i)= \mu$ when the sampling is done without replacement?

Can anyone please explain that?

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  • $\begingroup$ Imagine the population is a deck of cards; the value $1$ is assigned to an ace, the value $0$ to any other card; so $\mu=4/52=1/13.$ Now let $n=52,$ i.e., sample the whole deck without replacement. Let $X_i$ be the value of the $i$-th card. Do you believe that the expectations $E(X_1),E(X_2),\dots,E(X_n)$ are all equal? If not, which one do you think is the biggest? In other words: Suppose the whole deck is dealt out, face down; you want to draw an ace, and you can take any card you want. Which card would you pick, to maximize your chances of getting an ace? $\endgroup$ – bof Mar 21 '16 at 9:22
  • $\begingroup$ @bof: Okay, for the first draw, $\mathrm E(X_1)= \mu$ as all the 52 cards are available for selection. But what next? For the second draw, I've not 52 cards; so how can I expect $\mathrm E(X_2)= \mu\;?$ This is what I'm not understanding. $\endgroup$ – user142971 Mar 21 '16 at 9:27
  • $\begingroup$ Hmm. The 52 cards have been dealt out and are lying face down on the table. I can pick any card I want. If I pick the first card, I have one chance in 13 of drawing an ace. What if I pick the 17th card? Are my chances better or worse? Which card would you pick? What about the 52nd card? Is the ace more likely to be on the top or the bottom of the deck? $\endgroup$ – bof Mar 21 '16 at 9:31
  • $\begingroup$ The cards have been shuffled, and it's one chance in 13 that the top card is an ace. Suppose I cut the deck, with the result that the card that was originally on top is now the 27th card down. Is it now 1 chance in 13 that the 27th card is an ace? $\endgroup$ – bof Mar 21 '16 at 9:35
  • $\begingroup$ I believe the cause of your confusion is that you are thinking about conditional probabilities when you should be thinking of unconditional probabilities. $\endgroup$ – bof Mar 21 '16 at 9:37

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