Expected number of singlets breaking off from a chain

Question

In Feller's "An Introduction to Probability Theory and Its Applications", 3rd edition, volume 1, on page 226-227 he states the following:

Spores of the fungus Sordaria are produced in chains of eight. The chain may break into several parts, and ultimately the spores escape in projectiles from 1 to 8 spores. There are reasons to suppose that the breakages at the seven links are stochastically independent and that the links have the same probability p to break ... For Example, the spores located at the ends of the chain have probability p to become singlets whereas for all the other spores this probability equals $p^{2}$. By the addition rule therefore the expected number of singlets arising from one chain is given by $E_{1} = 2p + 6p^{2}$.

I understand the probability for each spore but I am struggling to see how he came to the expected value. My understanding of expectation was that we needed to multiply each possible value (1, 2, ..., 8) by its corresponding probability. I am confused how I would get the probabilities of each outcome though.

Thanks

Answer 1

Feller is using linearity of expectation (his "addition rule"): $\mathbb E[X+Y+Z]=\mathbb E[X]+\mathbb E[Y]+\mathbb E[Z]$

In this case let $X_i=1$ if spore $i$ is a singlet and $=0$ otherwise (and indicator random variable. You want the expected count of singlets $\mathbb E[X_1+X_2+\cdots+X_8]$ so need to find each $\mathbb E[X_i]$ so you can sum them. Your "multiply each possible value ... by its corresponding probability" will do this for you, but the possible values in each case are $0$ or $1$.

• For $i=1$ or $8$, you have $\mathbb E[X_i]=0 \times (1-p)+1\times p = p$
• for the other six $i$, you have $\mathbb E[X_i]=0 \times (1-p^2)+1\times p^2 = p^2$
• add these up and you get $2p+6p^2$, like Feller.

The distribution of the total number of singlets is not possible to calculate here without further assumptions, because you do not know the dependency of one chain link breaking on another breaking (you might guess that perhaps each chain link breaks independently of the others). But you do not need this to answer this question about the expectation.

---

If you did assume each of the $7$ chain links break independently with probability $p$, then with some effort it is possible to look at the $2^7=128$ possible patterns (not equally likely) and find the probabilities of the numbers of singlets:

• $0$ singlets with probability $(1-p)^7 + 5p(1-p)^6+6p^2(1-p)^5+p^3(1-p)^4$
• $1$ singlet with probability $2p(1-p)^6+12p^2(1-p)^5+12p^3(1-p)^4$
• $2$ singlets with probability $3p^2(1-p)^5+18p^3(1-p)^4+10p^4(1-p)^3$
• $3$ singlets with probability $4p^3(1-p)^4+20p^4(1-p)^3$
• $4$ singlets with probability $5p^4(1-p)^3+15p^5(1-p)^2$
• $5$ singlets with probability $6p^5(1-p)^2$
• $6$ singlets with probability $7p^6(1-p)$
• $7$ singlets with probability $0$
• $8$ singlets with probability $p^7$

You can then find the expected number of singlets as $\sum\limits_{n=0}^8 n\, p(n)$ which in this case with a lot of expansion and tidying up would also give $2p+6p^2$. It is not the easiest way to do it.