There's a question in the book "An Introduction to Mathematical Statistics and Its Applications" that I can't understand:
An urn contains $r$ red balls and w white balls. A sample of $n$ balls is drawn in order and without replacement. Let $X_{i}$ be $1$ if the ith draw is red and $0$ otherwise, $i = 1, 2,..., n$.
Show that $E(X_{i}) = E(X_{1}),i = 2, 3,..., n$.
How is that possible. At the 1th draw we have a probability of $\frac{r}{r+w}$ of getting a red ball. Thus $E[X_{1}]=1 \frac{r}{r+w}+0 \frac{w}{r+w}=\frac{r}{r+w}$ .
Thus for the 2th draw we have two possible cases:
- 1th draw we get a red ball with probability $\frac{r}{r+w}$.
- 1th draw we get a white ball with probability $\frac{w}{r+w}$.
Thus,
$$E[X_{2}]=\frac{r}{r+w}E[X_{2}|X_{1}=1]+\frac{w}{r+w}E[X_{2}|X_{1}=0]$$
$$=\frac{r}{r+w}[1\frac{r-1}{(r-1)+w}+0\frac{w}{(r-1)+w}]+\frac{w}{r+w}[1\frac{r}{(r-1)+w}+0\frac{w-1}{(r-1)+w}]$$
$$=\frac{r(r-1)}{(r+w)(r+w-1)}+\frac{wr}{(r+w)(r+w-1)}$$
$$=\frac{wr^{2}(r-1)}{(r+w)(r+w-1)}$$
Obviously $E[X_{1}] \not = E[X_{2}]$.
No, you have calculated the conditional probability for getting a red ball on draw 2 given that a red ball was obtained on draw 1.
$$\mathsf P(R_2\mid R_1)=\dfrac{r-1}{(r-1)+w}$$
While correctly evaluated that is not actually what you want. Rather you seek the marginal probability for the event.
The probability for getting a red ball on draw 2 is $r/(r+w)$. By the Law of Total Probability: $$\begin{align}\mathsf P(R_2)&=\mathsf P(R_2\mid R_1)\mathsf P(R_1)+\mathsf P(R_2\mid W_1)\mathsf P(W_1)\\&=\dfrac{r-1}{(r-1)+w}\cdot\dfrac{r}{r+w}+\dfrac{r}{r+(w-1)}\cdot\dfrac{w}{r+w}\\&=\dfrac{r}{r+w}\end{align}$$
And so on.
Per your edit, you should have the same result
$$\begin{align}\mathsf E(X_2)&=\mathsf E(X_2\mid X_1=1)~\mathsf P(X_1=1)+\mathsf E(X_2\mid X_1=0)~\mathsf P(X_1=0)\\&=\mathsf P(X_2=1\mid X_1=1)~\mathsf P(X_1=1)+\mathsf P(X_2=1\mid X_1=0)~\mathsf P(X_1=0) \\&=\dfrac{r-1}{(r-1)+w}\cdot\dfrac{r}{r+w}+\dfrac{r}{r+(w-1)}\cdot\dfrac{w}{r+w}\\&=\dfrac{r}{r+w}\cdot\dfrac{r+w-1}{r+w-1}\end{align}$$
In short, every ball has an equal chance of being in place $i$ in the draw, and $r$ of the $r+w$ are red.$$\mathsf P(R_i)=\dfrac{r}{r+w}$$
and $\mathsf E(X_i)=\mathsf P(R_i)$ ...