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Before, I begin I want say that the original question asks, "A couple decides to continue to have children until a daughter is born. What is the expected number of children until a daughter is born?"

Below, is the solution that they had in the back of the textbook (Statistical Inference, Second Edition, Roger L. Berger, George Casella) enter image description here What I do not understand is how to deduce summations that well.

First of all, how do they get $$\sum^\infty_{k=1}k(1-p )^{k-1}p = p - \sum^\infty_{k=1}\frac{d}{dp}(1-p )^{k}$$

Next, How do they simplify this as well $$-p\frac{d}{dp}{\bigg[\sum^\infty_{k=0}(1-p)^k-1\bigg]} = -p\frac{d}{dp}\bigg[\frac{1}{p}-1\bigg] = \frac{1}{p}$$

I would like this to be shown and explained intuitively because the statistics course I am taking requires me to be able to deduce summations on my own, and it seems to be a huge part of understanding the subject. Thank you ahead of time.

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  • $\begingroup$ the answer is: you see that, if you've practiced enough :( sorry! It's really like that. For the first question, there might be other ways to arrive at $\frac1p$, but the path they chose here is "I see a term to the power of something minus one, that times that something – that looks like a derivative", and that's it. It's an Ansatz they tried and that worked. Note that this has exactly nothing to do with your question's title! $\endgroup$ – Marcus Müller Jul 21 at 19:05
  • $\begingroup$ It is just crazy how I am expected to know how to answer that. I found the solution online, but my instructor just wanted me to answer the question and show my work. I do not know how I am expected to do that when this is my first stats class :/ $\endgroup$ – Alexander Martinez Jul 21 at 19:10
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    $\begingroup$ the best advice is tp grab a calculus text book and look at the chapter on Taylor series and practice. If you aren’t a little familiar with doing manipulations yourself it’ll be hard to see them in the wild. Geometric series and integrating/differentiating them are especially useful $\endgroup$ – Nap D. Lover Jul 21 at 19:11
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    $\begingroup$ your not expected to know how to answer this from your first stats class alone; this is usually exercised, a lot, in "higher maths for XYZ" or "calculus" courses that you have before stats. $\endgroup$ – Marcus Müller Jul 21 at 19:11
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As an alternate method that does not require the use of Geometric Series:

Let $E$ denote the answer, then consider the possible outcomes from the first child. Either the child is a daughter ($p$) in which case the answer is $1$ or it is a son ($1-p$) and the answer is now $1+E$.

Thus $$E=p\times 1 + (1-p)\times (E+1)\implies E=\frac 1p$$

In general, the use of Geometric Series only makes sense in those situations (as here) in which one can easily enumerate all the possible outcomes. Generally speaking, however, this is not possible.

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This is a long-winded comment that is intended as a supplement to the other already posted responses. This is not an answer.

The foundation of the geometric series is the idea that for any real number $x \neq 1$, and any positive integer $n$, you have that

$$\left(1 + x + x^2 + \cdots + x^n\right) \times (1 - x) = \left[1 - x^{(n+1)}\right]. \tag1 $$

Formulas around geometric series are all based on equation (1), above.

For example, from equation (1), you have that

$$\left(1 + x + x^2 + \cdots + x^n\right) = \frac{1 - x^{(n+1)}}{1 - x}. \tag2 $$

Now, for $~x \neq 1, ~n \in \Bbb{Z^+},~$ let $E_n(x)$ denote $\displaystyle \frac{x^{(n+1)}}{1 - x}$.

Then you have that

$$\left(1 + x + x^2 + \cdots + x^n\right) = \frac{1}{1 - x} - E_n(x). \tag3 $$

This means that when $\displaystyle \left(1 + x + x^2 + \cdots + x^n\right)$

is estimated as $\displaystyle \frac{1}{1-x}$, then $E_n(x)$ is the error term in the estimation.

Therefore, a natural question to ask is :
as $n \to \infty$, for which values of $x$ will $E_n(x)$ go to $(0)$?
[Denote this question as Q1].

Note that since fixed values of $x \neq 1$ are being considered, question Q1 is equivalent to asking: as $n \to \infty$, for which values of $x$ will $x^{(n+1)}$ go to $(0)$?
[Denote this question as Q2].

It is an intermediate result in Real Analysis (AKA Calculus) that the answer to question Q2 is that
as $n \to \infty, ~x^{(n+1)} \to (0) \iff |x| < 1.$

What this implies is that for any $|x| < 1,$ the error term in equation (3) above will go to $(0)$ as $n \to \infty.$

This implies that for all $|x| < 1,$ you have that $\displaystyle \lim_{n \to \infty} \sum_{k=0}^n x^k = \frac{1}{1-x}.$

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First of all, how do they get $$\sum^\infty_{k=1}k(1-p )^{k-1}p = p - \sum^\infty_{k=1}\frac{d}{dp}(1-p )^{k}$$

By way of a typo. That is not what they should get.

$$\begin{align}\sum^\infty_{k=1}k(1-p )^{k-1}p ~&=~ p\sum_{k=1}^\infty k(1-p)^{k-1}\\ &=~-p\sum_{k=1}^\infty \dfrac{\mathrm d (1-p)^k}{\mathrm d p}\\[3ex]&=~-p\dfrac{\mathrm d ~~}{\mathrm d p}\left(\sum_{k=1}^\infty(1-p)^k\right)\\&=~-p\dfrac{\mathrm d ~~}{\mathrm d p}\left(\sum_{k=0}^\infty(1-p)^k-(1-p)^0\right)\\&=~-p\dfrac{\mathrm d ~~}{\mathrm d p}\left(\dfrac 1p-1\right)\\&=~-p\left(-\dfrac 1{p^2}\right)\\&=\dfrac 1p \end{align}$$

Next, How do they simplify this as well $$-p\frac{d}{dp}{\bigg[\sum^\infty_{k=0}(1-p)^k-1\bigg]} = -p\frac{d}{dp}\bigg[\frac{1}{p}-1\bigg] = \frac{1}{p}$$

By way of change of variables. When $\lvert 1-p\rvert \leqslant 1$ (which it is because $p$ is a probability measure), then the series converges and so we have:

$$\begin{align}\sum_{k=0}^\infty (1-p)^k~&=\sum_{j=0}^\infty (1-p)^j\\&=~(1-p)^0+\sum_{j=1}^\infty (1-p)^j\\&=~1+\sum_{k=0}^\infty (1-p)^{k+1}\\&=~1+(1-p)\sum_{k=0}^\infty (1-p)^k\\\therefore\quad(1-(1-p))\sum_{k=0}^\infty (1-p)^k~&=~1\\\sum_{k=0}^\infty(1-p)^k~&=~\dfrac 1p\end{align}$$

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