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I'm studying for my discrete exam and I can't figure out this problem in the review, any help is appreciated.

When Jane and Bob have a child, this child is a boy with probability 1/2 and a girl with probability 1/2, independent of the gender of previous children. Jane and Bob stop having children as soon as they have a girl.

What are the random variables:

B = the number of boys that Jane and Bob have

G = the number of girls that Jane and Bob have.

What are the expected values E(B) and E(G)

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  • $\begingroup$ Do you know the mean of a geometrically distributed random variable or do you need to derive it? $\endgroup$ – André Nicolas Dec 4 '14 at 0:07
  • $\begingroup$ no, i think i need to derive it $\endgroup$ – IamGrateful Dec 4 '14 at 0:11
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The mean number of girls is obviously $1$.

For the mean number of boys, call having a boy a failure. We want to find the expected number of failures until the first success. We could do it by summing an infinite series, or by conditioning, or by thinking.

Let $a$ be the expected number of failures. If there is a success on the first trial (probability $1/2$), then the number of failures is $0$. If there is a failure on the first trial, the conditional expectation of the number of additional failures, given there was a failure on the first trial, is $a$. Thus in that case the expected total number of failures is $1+a$. Thus $$a=(1/2)(0)+(1/2)(1+a).$$ Solve for $a$. We get $a=1$. The expected number of boys is also $1$.

If we think about it, the result is obvious. On any birth, the expected number of boys is the same as the expected number of girls.

Remarks: $1.$ With the series approach, we end up summing the infinite series $$\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\cdots.$$ There are arious methods for summing this sort of series. The problem has been solved repeatedly on MSE.

$2.$ If you already know that the geometric with parameter $p$ has expectation $\frac{1}{p}$, then you can see that the expected total number of children is $2$, and hence the expected number of boys is $1$.

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  • $\begingroup$ How did u solve for a. that u get a = 1. I got a = -1? $\endgroup$ – Need Help Dec 4 '14 at 0:42
  • $\begingroup$ Steps: $a=\frac{1}{2}+\frac{1}{2}a$. So $\frac{1}{2}a=\frac{1}{2}$. So $a=1$. Or else multiply through by $2$. We get $2a=1+a$ so $a=1$. $\endgroup$ – André Nicolas Dec 4 '14 at 0:52

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