
#1




Joint PMFs
Hello. I have been struggling on this question.
Let X and Y be independent Geo(1/3) random variables. Let V =min(X,Y) and let W be defined as follows: W = 0 if X< Y, W=1 if X=Y W=2 if X > Y. Let p{V,W}(v,w) be the joint probability mass function of V and W. Compute p{V,W}(2,0). I know that I have to make a pmf of V and W in terms of X and Y, but I am stuck on how to do so. Also, how would I use the geometric random variable pmf into this problem? If someone can get me started on this problem, that would be appreciated. 
#2




Quote:
To find the probability that X=2 and Y>2, that is the Pr(X=2)*Pr(Y>2). For Pr(X=2), plug into the Geometric distribution: (11/3)^(21) * (1/3) = 2/9. Now calculate Pr(Y>2) as the complement 1P(Y<=2)....there are three options to consider. Multiply that result with 2/9.
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#4




The problem alone doesn't tell you whether geometric is counting failures or trials here. Your answer of is correct if you're counting trials. If you're counting failures it would be .
Also just for fun and in case you find this useful (all true whether counting failures or trials): The PMF of is is Geometric with parameter In general if and are independent and geometric with parameters and respectively, then is geometric with parameter . Last edited by Z3ta; 11202017 at 11:28 AM.. 
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