Question

How can i complete finding the conditional probability function? (details inside)

we toss a regular coin 50 times:

X - the number of heads from 50 tosses.

Y - the number of heads from the first 20 tosses.

1) find the conditional probability function of Y if given that X=i for {i=0,...,50}, and the conditional probability function of X if given Y=j so that {j=0,...,20}

2) calculate `E[X∣Y=j]`

so i tried to connect X and Y by another variable, `Z=X-Y`, and then to calculate `P_{X|Y}(x | y) =\frac{P_{XY}{(x,y)}}{p_Y(y)}` and `P_{Y|X}(y| x) =\frac{P_{XY}{(x,y)}}{p_X(X)}`. to calculate `P_{X,Y}(x,y)` i used `P_{X,Y}(x,y)=P_{Z}(x-y)\cdot P_Y(y)` - but i don't know how to continue from here.

for2) `E[X∣Y=j]` : `E(X|Y=j)= E(Z+Y| Y=j)` but i don't know how to continue. would really appreciate your help with it.

thank you very much for your help

Answer 1

`"1a) "P("Y=k|X=i") = { ((C(i,k) C(50-i, 20-k)) / (C(50,20)), if k<= i), (0, if k>i) :}`

`"1b) "P("X=k|Y=j") = { ((C(30, k-j)) / 2^30, if k >= j), (0, if k < j) :}`

`"2) "E[ X |Y = j ] = j + 15`

Explanation:

`"1a) "`
`P("Y=k|X=i") = 0, if k > i`
`P("Y=k|X=i") = (C(i,k)*20*19*...*(21-k)*30*29*...*(31-(i-k)))/(50*49*...*(51-i)), if k <= i`
`= C(i,k) (((20!)/((20-k)!)) ((30!)/((30-i+k)!))) / ((50!)/((50-i)!)), if k <= i`
`= { ((C(i,k) C(50-i, 20-k)) / (C(50,20)), if k<= i), (0, if k>i) :}`

`"1b) "`
`P("X=k|Y=j") = 0, if k < j`
`P("X=k|Y=j") = P("k-j times head in the last 30 tosses")`
`= { ((C(30, k-j)) / 2^30, if k >= j), (0, if k < j) :}`

`"2) "`
`E[ X | Y = j ] = (j*C(30,0) + (j+1)*C(30,1) + ... + (j+30)*C(30,30))/2^30`
`= j + (C(30,1) + 2*C(30,2) + ... + 30*C(30,30))/2^30`
`= j + (30*(1 + C(30,1) + ... + C(30,14)) + 15*C(30,15))/2^30`
`= j + (30*(2^30/2 - (C(30,15))/2) + 15*C(30,15))/2^30`
`= j + (15*(2^30 - C(30,15)) + 15*C(30,15))/2^30`
`= j + 15`

`"This is a very logical result as we certainly have j head"`
`"tosses and can expect half of the remaining 30 tosses"`
`"to give head on average."`