The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

This discussion is now closed.

Check out other Related discussions

Probability Question

OK, I was going to put this in the University Maths forum, but it appears that it's not welcome there. Anyway...

X and Y are integer-valued random variables with joint mass function
P(X=k,Y=m) = q^(k+m) * p^2

For k,m >= 0, where p+q=1 and 0<p<1

Are X and Y independent (+ why?)

Find:
- the marginal distributions of X and Y
- P(X+Y = n)
- P(X=k|X+Y=n)
Reply 1
Avatar for kikzen
kikzen
14
q = 1-p
- the marginal distributions of X and Y

sum(mE{0,1,2,...} of P(X=k;Y=m) )
=p^2.(1-p)^k sum(m=0 to infinity of (1-p)^m )
= p(1-p)^k

similarly for y
Reply 2
Avatar for kikzen
kikzen
14
for
- P(X+Y = n)

i think you use that

P(X+Y = n) = sum in i(P(X=i)P(Y=n-i) )
using the marginals you found before
Reply 3
Avatar for Jonny W
Jonny W
8
X and Y are independent because P(X = k, Y = m) factorizes into (a function of k)*(a function of m):

P(X = k, Y = m)
= [p*q^k] * [p*q^m]

I chose that factorization because each of the terms in square brackets is a probability mass function.

P(X = k)
= (sum over m) P(X = k, Y = m)
= p*q^k

P(Y = m)
= (sum over k) P(X = k, Y = m)
= p*q^m

P(X + Y = n)
= (sum over m with 0 <= m <= n) P(X = m, Y = n - m)
= (sum over m with 0 <= m <= n) q^n * p^2
= (n + 1) q^n * p^2

P(X = k | X + Y = n)
= P(X = k, X + Y = n) / P(X + Y = n)
= P(X = k, Y = n - k) / P(X + Y = n)
= q^n * p^2 / [(n + 1) q^n * p^2]
= 1/(n + 1) . . . . . for 0 <= k <= n

Ie, given that X + Y = n, X is uniformly distributed over {0, 1, ..., n}.
Reply 4
Avatar for flyinghorse
flyinghorse
OP
Thanks both :smile:
Reply 5
Avatar for flyinghorse
flyinghorse
OP
Just another little query:
how would I find E(X|X+Y=n) ? Just use P(X|X+Y=n) and do it the normal way (ie the same way as just finding E(X))?

Related discussions

Latest

Trending

Trending

Articles for you