# Probability calculation problem for a ranked set sample

#### Blain Waan

##### New Member
Let us consider a case where a ranked set sampling scheme from $n^2$ subjects has been adopted. There are a total of $n$ RSS blocks. So each RSS block contains $n$ subjects. We will follow up them. Some of these subjects will fulfill a certain criterion and get a treatment and some will not be able to fulfill the criterion and will not get the treatment. Again the person who will be able to complete the criterion first in the 1st RSS block, who will be able to fulfill the criterion second in the 2nd RSS block $\ldots$ and last in the last RSS block will receive a "special treatment". So, if all fulfilled the criterion then we would expect $n$ subjects to receive the "special treatment". But we have observed that out of $n^2$ subjects $M$ patients do not fulfill the criterion hence receive no treatment.

The response times of the other $n^2-M$ subjects are known to us. We also know the lifetimes for $M$ non-responders. And it can happen that a subject dies before responding to the criterion. Now what should be the probability that the $i$-th subject gets the special treatment?

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**The way I tried:**

**1)** First of all the probability of getting the **special treatment** for those who do not fulfill the criterion is $0$.

**2)** If we order the response times of the $n^2-M$ patients, then say the response times are $x_1<x_2<\ldots<x_i<\ldots<x_{n^2-M}$. So for $i$ to be the $k$-th smallest in a RSS block $(k-1)$ $x$'s should come before $x_i$ from the $x$'s. Let, among the $M$ non-responders, $M_i$ were observed to die before $x_i$ remitted. There can also be lowest $0$ to highest $m=min\{(n-k),M_i\}$ elements from these $M_i$ non-responders before $x_i$.

After $x_i$, there will be $(n-k-m)$ elements from the rest of $(n^2-i-M_i)$ elements.

I want to find out the probability that the $i$-th subject gets the **special treatment**.

#### BGM

##### TS Contributor
Let say you have the distribution of the response time $$X_i$$ and the life time $$T_i$$ (maybe capped by the maximum experimental observation time) for the individuals.

First of all do you know if a particular $$i$$-th individual belongs to a specific block?

Let say given the $$i$$-th individual belongs to the $$k$$-th block. If I have understand your given information correctly, he gets a special treatment if he is the $$k$$-th responder in the block.

So now you need to condition on the number of responses in that block:

$$R_k = \sum_{i=1}^n \mathbf{1}\{X_i < T_i\}$$

If we have the nice i.i.d. assumptions on $$(X_i, T_i)$$, then $$R_k \sim \text{Binomial}(n, p = \Pr\{X_i < T_i\})$$

Note the individual has the possibility of obtaining the special treatment only when $$R_k \geq k$$.

When this happen, by the i.i.d. assumption the probability of the individual actually responded, i.e. $$X_i < T_i$$, given $$R_k = r$$ is just $$\frac {r} {n}$$

Also the rank of a particular individual in the responded group is just follow a discrete uniform distribution, i.e. the probability of being exactly having the $$k$$-th rank given $$R_k = r$$ is $$\frac {1} {r}$$.

Combining these things together, by Law of Total probability, the probability of a particular individual receiving the special treatment given he is located at the $$k$$-th block is

$$\sum_{r=k}^n \frac {1} {r} \frac {r} {n} \binom {n} {r} p^r (1 - p)^{n-r} = \frac {1} {n} \Pr\{R_k \geq k\}$$

Last but not least if the individual is uniformly selected over the $$n$$ possible blocks, then each block is equally-likely to be selected and thus the probability of belonging to a particular block will just be $$\frac {1} {n}$$ and by Law of Total Probability again, the unconditional probability will just be

$$\sum_{k=1}^n \frac {1} {n} \frac {1} {n} \Pr\{R_k \geq k\} = \frac {1} {n^2} \sum_{k=1}^n \Pr\{R_k \geq k\}$$

If the distribution among the blocks again is the same, the above summation will just be $$E[R_k]$$ and therefore the whole expression reduced to

$$\frac {1} {n^2} np = \frac {1} {n} \Pr\{X_i < T_i\}$$