# For a sequence of experiments where each $X$ is the number of trials until success with varying $p$, is each $X$ independent?

Assume that, every time you buy a box of Wheaties, you receive a picture of one of the $$n$$ baseball player. Let $$X_k$$ be the number of additional boxes you have to buy, after you have obtained $$k-1$$ different pictures, in order to obtain the next new picture. Thus $$X_1 = 1$$, $$X_2$$ is the number of boxes bought after this to obtain a picture different from the first pictured obtained, and so forth.

If I want to find the variance for the number of boxes before getting half of the players' pictures (assume there are $$2n$$ players), my book states the following:

$$p_{X_k} = \frac{2n-k+1}{2n}$$

Since this is a geometric distribution, $$V(X)=\frac{1-p}{p^2}$$, so

$$V(X_k) = \frac{2n(k-1)}{(2n-k+1)^2}$$

So variance for the total number of boxes before getting the first half of the players' pictures is:

$$\sum_{k=1}^{13}\frac{26(k-1)}{(26-k+1)^2}$$

For expected value I get that regardless of whether $$X_1,X_2..,X_k$$ are independent $$E(X_1+X_2+..+X_k)=E(X_1)+E(X_2)...+E(X_k)$$ but I don't think this is true for variance. Based on the above solution it seems like that the number of boxes to get the kth players' picture (random variables $$X_1,X_2..X_k$$) mutually independent? If so, is there a way to show this? This isn't intuitively clear to me, since $$p_k$$ is variable and dependent on $$k$$.

Variables $$X_1,X_2,...X_k$$ are indeed mutually independent.
Of course $$X_1$$ = $$1$$, lets say you received Lebrons picture.
Imagine that $$X_2$$ = $$1000$$. So it means that you bought $$999$$ cards and for all of them you received Lebrons picture. Only on try $$1000$$ you received a new picture.
Does it somehow affect the $$X_3$$? the fact that you had to buy $$1001$$ boxes to obtain two different pictures doesnt change the probability of obtaining new card. All it matters for $$X_3$$ is that you have only two different pictures. So is for any $$X_i$$
• I think what confused me was that $p$ was variable in this case, but $p_i$ is specific to an $X_i$ and is dependent only on the position of $X_i$ in the sequence of $X$'s, which has nothing to do with an event in the experiment ({X_1=a_1,...X_{i-1}=a_{i-1}}) for which $X_i$ is conditioned on? – Yandle May 20 at 17:52
• $p$ is not variable. It is a function of index $i$. Once you know how many different cards you got - "$i-1$", you can calculate $p_i$. – Markoff Chainz May 20 at 18:21