Poisson distribution

[Rup]

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]

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]

Someone posted with this exact same problem a few days back

http://www.actuary.com/actuarial-discussion-forum/showthread.php?t=1362

[mallkins]

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]

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]

I think I got my answer. It is making sense.Thank you all!