Question

Suppose that Y1,...,Yn is a random sample from Bernoulli(p). Find the distributions of Y(1)=min{y1,...,yn} and Y(n)=max{y1,...,yn}.

Answer 1

order statistics... look it up

Answer 2

Wow, so helpful.

Answer 3

If I knew what do with it, I wouldn't be here....

Answer 4

well, since it's bernoulli, it may simplify a bit. bernoulli is 1 with probability p. so the y's will either be 0's or 1's.
Are the bernoulli's iid?

Answer 5

and its' the \[y_i\] which are bernoulli, yeah? not the \[Y_i\]

Answer 6

capital Y's throughout.

Answer 7

what's iid?

Answer 8

independent, identically distributed

Answer 9

ah yes, they are. Bernoulli's are defined as independent. and their dist. is identical as stated in the problem.

Answer 10

so Y_(1) can either be 0 or 1.

it can be 1 iff all of the bernoulli's are 1, ow it's 0.

P(Y_(1)=1)=P(all Y's are 1) = p^n
=> P(Y_(1)=0)=1-p^n

Answer 11

for the max, you approach the other way round...

P(Y_(n)=0) = P(all of the Y's are 0) = (1-p)^n

P(Y_(n) = 1) = 1-(1-p)^n

Answer 12

So they're both Bernoulli with a different p. For Y_(1) it's Bernoulli (p^n) and for Y_(n) it's Bernoulli [(1-p)^n]

Answer 13

oops, i mean 1-(1-p)^n for Y_(n)

Answer 14

does that make sense?

Answer 15

Kind of

Answer 16

let's focus on the min...\[Y_{(1)}\] it's either going to be 0 or 1, right?

Answer 17

right

Answer 18

so, how can it be 1? in other words, what has to happen in order for it to be 1?

Answer 19

all would have to be 1

Answer 20

thus the probability of that is p^n due to the independence

Answer 21

right and since they're iid, =>\[ P(\text{all }Y_i \text{'s} = 1) = p \cdot p \cdot p \cdot (\cdots) \cdot p \,\,\text{(n \times)} = p^n\]

Answer 22

but if the min is = 0?

Answer 23

then since \[Y_{(1)}\] can only be 0 or 1, \[P(Y_{(1)}=0)=1-p^n\]

Answer 24

thus, \[Y_{(1)}\sim\text{Bernoulli }(p^n)\]

Answer 25

It's Bernoulli(p^n) because...?

Answer 26

it can only be 0 or 1, right?

Answer 27

yep and it's 1("success") with probability p^n

Answer 28

got it.

Answer 29

so now go over the max... and make sure you get that one too.

Answer 30

good question... made me have to go into the deep recesses of my memory!

Answer 31

it's Bernoulli with p =1-(1-p)^n

Answer 32

yep, you understand how we got that?

Answer 33

yeah, similar to the MIN, but start with Y=0

Answer 34

you got it... all Y_i's are 0 in order for Y_(n) to be 0

Answer 35

=>P( Y_(n)=0) = (1-p)^n thus P(Y_(n)=1) = 1 - (1-p)^n ("success")

Answer 36

Thanks!

Answer 37

have a look at this for other order stats...

Answer 38

thanks for the question!