# Math Help - problem of random variables

1. ## problem of random variables

Suppose X1;X2 are two independent geometrically
distributed random variables whose parameters are respectively 1/2 and 1/3 : What
is P(X1 = X2)?

i dont know what are the parameters refered to ?

2. Originally Posted by kin
Suppose X1;X2 are two independent geometrically
distributed random variables whose parameters are respectively 1/2 and 1/3 : What
is P(X1 = X2)?

i dont know what are the parameters refered to ?
The parameters are the probability of success for each random variable, that is, the value of p.

3. Originally Posted by kin
Suppose X1;X2 are two independent geometrically
distributed random variables whose parameters are respectively 1/2 and 1/3 : What
is P(X1 = X2)?
It seems to me that P(X1 = X2)= sum(x=0..+oo) P(X1 = x AND X2 = x)
Since X1 and X2 are i.i.d, P(X1 = x AND X2 = x) = P(X1 = x) P(X2 = x)
Now, substituting the expressions for each probability above you should be done

4. Originally Posted by tossifan
It seems to me that P(X1 = X2)= sum(x=0..+oo) P(X1 = x AND X2 = x)
Since X1 and X2 are i.i.d, P(X1 = x AND X2 = x) = P(X1 = x) P(X2 = x)
Now, substituting the expressions for each probability above you should be done
summation(x=0..+oo) [P(X1 = x AND X2 = x)]

=p(x1=0,x2=0)+ p(x1=1, x2=2)+p(x1=3,x2=3).....

=p(x1=0)p(x2=0)+ p(x1=1)p(x2=1)+......

where p(x1=x)=(1/2)( (1-1/2)^x);p(x2=x)= (1/3)( (1-1/3)^x)<--am i right here??

5. Originally Posted by kin
summation(x=0..+oo) [P(X1 = x AND X2 = x)]

=p(x1=0,x2=0)+ p(x1=1, x2=2)+p(x1=3,x2=3).....
=p(x1=0,x2=0)+ p(x1=2, x2=2)+p(x1=3,x2=3).....

Originally Posted by kin
=p(x1=0)p(x2=0)+ p(x1=1)p(x2=1)+......

where p(x1=x)=(1/2)( (1-1/2)^x);p(x2=x)= (1/3)( (1-1/3)^x)<--am i right here??
No, it should be P(X2=x)=(1/3) (1-1/3)^(x-1)

Give a look at the definition of geometric distribution.

6. Originally Posted by tossifan
=p(x1=0,x2=0)+ p(x1=2, x2=2)+p(x1=3,x2=3).....

No, it should be P(X2=x)=(1/3) (1-1/3)^(x-1)

Give a look at the definition of geometric distribution.
i start from x2= 0 so that P(X2=x)=(1/3) (1-1/3)^x;

if i start from x2=1, then i should be P(X2=x)=(1/3) (1-1/3)^(x-1),

are they not equivalent..?