Coupon collecting with unequal probabilities (S.Ross Min-Max approach)

Question

While going through a method to solve the coupon collecting problem , the author posed the problem as follows :

Suppose there are $n$ types of coupons and that each time one collects a coupon, it is, independently of previous coupons collected, a type $i$ coupon with probability $~~p_i~$,$~~\sum_{i=1}^n p_{i}=1$ .

Find the expected number of coupons one needs to collect to obtain a

complete set of at least one of each type.

Then in his attempt to the solution he defined his random variables the following way :

let $X_{i}$ denote the number of coupons one needs to collect to obtain a type $i$ coupon , and let $X$ be the number of coupons one needs to collect to obtain a complete set of at least one of each type.

After that he made this proposition : $~~~~X= \underset{i=1,...n}{{max}} {X_{i}}$ .

Question : Why is this formula true ? To me this relation doesn't sound obvious at all , could someone help me clarify that please ?

Source : A first Course In Probability (S.Ross ) p320.

Answer 1

$X_i$ is not, as the other answer states, the expected number of draws to obtain coupon $i$. It is the number of draws to obtain coupon $i$. It is a random variable, not the expected value of that random variable. Likewise, $X$ is a random variable, namely the number of draws required to obtain at least one coupon of each type; it is not the expected value of that number. The equation

$$ X= \underset{i=1,\ldots,n}\max {X_{i}} $$

is true for the random variables but not for their expected values. For the random variables, it is true simply because you've collected each coupon type at least once precisely at the time at which you've collected the last coupon type at least once.

Answer 2

I believe the point being made is this:

Remember $X_i$ is the number of draws until coupon $i$ is finally picked. Therefore, $\text{max}(X_i)$ represents the most draws needed to obtain a certain coupon. In theory if you finally obtain this "rare" coupon, you should have drawn at least one of all the others.

However, note that the expected value of $X$ is not equal to $\text{max}(X_i)$