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Probability 20 Online
OpenStudy (anonymous):

let X1 and X2 be iid possion variable with mean 1. Then P[max( X1,X2)] is?

OpenStudy (anonymous):

What is the question? Sorry, I'm having trouble reading it! :P

OpenStudy (anonymous):

Let \(X=\max(X_1,X_2)\), then the cdf of \(Y\) is given by \[\begin{align*}F_X(x)&=P(X_1\le x,X_2\le x)\\ &=P(X_1\le x)P(X_2\le x)&\text{since }X_1\text{ and }X_2\text{ are i.i.d.}\\ &=F(x)F(x)&\text{by definition of c.d.f.}\\ &=\left[F(x)\right]^2 \end{align*}\] \(P(X)=P(\max(X_1,X_2))\) is then given by the pdf of \(X\): \[f_X(x)=\frac{d}{dx}\left[F(x)\right]^2=2F(x)f(x)\] So really, the question is a matter of multiplying the cdf and pdf of the Poisson distribution. I don't know them off the top of my head.

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