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Rup
February 15th 2006, 05:14 AM
Let X1 and X2 form a random sample of a Poisson distribution. The Poisson distribution has a mean of 1. If Y= min(X1,X2), then P(Y=1)=? Ans: (2e-3)/e^2

The first step is
P(Y=1)= P(X1=1)intersection(X2>=1) + P(X2=1)intersection(X1>=2)

I am not really getting X1 >=2

Can anybody explain?

Thanks

Ken
February 15th 2006, 07:07 AM
Not sure what your question is. Are you not sure why it's X1>=2 or not sure how to calculate it?

If it was X1>=1, you would be double counting the situation where (X,Y)=(1,1)

P(X1>=2)=1-P(X=0)-P(X=1)

amark82
February 15th 2006, 10:09 AM
Someone posted with this exact same problem a few days back

mallkins
February 15th 2006, 02:38 PM
Let X1 and X2 form a random sample of a Poisson distribution. The Poisson distribution has a mean of 1. If Y= min(X1,X2), then P(Y=1)=? Ans: (2e-3)/e^2

The first step is
P(Y=1)= P(X1=1)intersection(X2>=1) + P(X2=1)intersection(X1>=2)

I am not really getting X1 >=2

Can anybody explain?

Thanks

In the original question, is it stating that X1 and X2 are both greater than zero? If not then surely in the P(Y=1)= P(X1=1)intersection(X2>=1) + P(X2=1)intersection(X1>=2) the X1 and X2 equal to zero are not accounted for.

Ken
February 15th 2006, 02:56 PM
In the original question, is it stating that X1 and X2 are both greater than zero? If not then surely in the P(Y=1)= P(X1=1)intersection(X2>=1) + P(X2=1)intersection(X1>=2) the X1 and X2 equal to zero are not accounted for.

The problem states Y = min(X1, X2) so for Y to be equal to 1, X1 and X2 have to be both greater than zero.

Rup
February 15th 2006, 05:03 PM
I think I got my answer. It is making sense.Thank you all!