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An integer is repeatedly drawn at random from $1, 2, . . . , 10$. What are the expected value and the standard deviation of the number of integers from $1, 2, . . . , 10$ that do not show up in $20$ drawings?

Let $X_i$ be the random variable that assumes value $1$ if the number $i$ doesn't show up in $20$ drawings and $0$ otherwise. So $\mathbb{P}(X_i=1)=(\frac{9}{10})^{20}$. Since $\mathbb{E}[X_i]=0\cdot (\frac{1}{10})^{20}+1\cdot (\frac{9}{10})^{20}=(\frac{9}{10})^{20}$, I know that:

  • $\mathbb{E}[X]=\mathbb{E}[X_1]+...+\mathbb{E}[X_{10}]=(\frac{9}{10})^{20}+...+(\frac{9}{10})^{20}=10\cdot (\frac{9}{10})^{20}=1,216$

  • $\operatorname{Var}[X]=\mathbb{E}[X^2]-\mathbb{E}[X]^2=\space{?}-(1,216)^2$

How do I find $\mathbb{E}[X^2]$?

EDIT:

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Francesco Totti
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  • Each $X_i$ looks like a binomial distribution. Can you use the binomial moment-generating function? – Raiyan Chowdhury Sep 13 '20 at 16:17
  • @RaiyanChowdhury No, the exercise falls within the first section of program (in which the FGM has not yet been introduced). – Francesco Totti Sep 13 '20 at 16:23
  • If $X$ out of the $10$ integers don't show up after the $20$ drawings then $$X \sim \text{Binomial}\Big[10,(9/10)^{20}\Big]$$ – Matthew H. Sep 13 '20 at 16:26
  • @RaiyanChowdhury Maybe I have to use the same formula written here. https://math.stackexchange.com/questions/3824636/variance-of-x-if-x-i-sim-operatornameber-fracr-1rm?noredirect=1#comment7887564_3824636 – Francesco Totti Sep 13 '20 at 16:27
  • @MatthewHolder So $\operatorname{Var}[X]=np(1-p)=10\cdot ((9/10)^{20})\cdot (1-((9/10)^{20})$? – Francesco Totti Sep 13 '20 at 16:31
  • Yes. You can just use the formula for the variance of a binomial random variable. I am writing a solution for you to indicate how to formalize this a bit. – Matthew H. Sep 13 '20 at 16:32
  • @MatthewHolder However in this way I obtain $\operatorname{Var}[X]=1,0679\Rightarrow \sigma(X)=\sqrt{1,0679}=1,0334$, and the result is $0,88$. – Francesco Totti Sep 13 '20 at 16:35
  • Hmm. Okay. Let me think about this a little bit more. – Matthew H. Sep 13 '20 at 16:38

1 Answers1

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We have $$\mathsf E(X^2)=\sum_{i,j=1}^{10} \mathsf E(X_i X_j)=10\mathsf E(X_1^2)+2\binom{10}2\mathsf E(X_1X_2).$$

Now, $$\mathsf P(X_1X_2=1)=\mathsf P(\text{$1$ and $2$ don't appear in any of the drawings})=\left(\frac8{10}\right)^{20}=1~-~\mathsf P(X_1X_2=0).$$ Can you finish it from here?

EDIT: There was a typo (it should be $10\mathsf E(X_1^2)$, not $10\mathsf E(X_1)$.)

Maximilian Janisch
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  • Ooooh, I was right then. It seems that you used the same formula here: https://math.stackexchange.com/questions/3824636/variance-of-x-if-x-i-sim-operatornameber-fracr-1rm?noredirect=1#comment7887564_3824636. Could you please explain me why you have written $\mathbb{E}[X^2]$ in that way? – Francesco Totti Sep 13 '20 at 16:50
  • @FrancescoTotti FYI this would yield the same result as what Matthew Holder wrote if you work out the math, since $\sum X_i \sim \text{Binomial}(10, (9/10)^{20})$. The sum of independent Bernoullis is a Binomial. – Clarinetist Sep 13 '20 at 17:00
  • @Clarinetist Please, see my edit. – Francesco Totti Sep 13 '20 at 17:03
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    The $X_1, \dots , X_{10}$ aren't independent. If $X_j$ represents the number of times the integer $j$ shows up then $P(X_1=5|X_2=16)=0$. – Matthew H. Sep 13 '20 at 17:03
  • @MatthewHolder Ugh, I knew there had to be some sort of dependence structure I'm missing here. I'll try to give this a shot. – Clarinetist Sep 13 '20 at 17:05
  • We need to find $P(X_j =0)$ for a given $j\in [10]$. If we can do that then the rest is history. I tried to employ a "stars and bars" technique, but it's not getting me anywhere. For example, I was thinking of counting the number of ways we could get $X_1 =0$ by dividing the numbers of non$-$negative solutions to $X_2 + \dots +X_{10}=20$ by the number of non$-$negative solutions to $X_1 + \dots + X_{10}=20$ and saying $P(X_1 =0)=\frac{{28 \choose 8}}{{29 \choose 9}}$ but it's not looking too promising. – Matthew H. Sep 13 '20 at 17:09
  • @MatthewHolder But $X_1,X_2$ can assume only the values $1$ and $0$. With $X_2=16$ you're saying that the number $2$ shows up $16$ times? – Francesco Totti Sep 13 '20 at 17:12
  • Yes. I defined my $X_j$ variables differently than you did. – Matthew H. Sep 13 '20 at 17:13
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    @Francesco I just expanded the square of a sum. For instance, try expanding $(a+b+c+d)^2$ for any real numbers $a,b,c,d$. – Maximilian Janisch Sep 13 '20 at 17:18