# Find $E(X)$ and $E(Y)$ and $\operatorname{Cov}(X,Y)$

Consider a sequence of Bernoulli trials with the probability of success $p$. Suppose you started the game with a run of successes followed by the run of failures (note that you can learn that unlucky run is over if and only if it is followed by a success). Let the random variable $X$ be the number of successful trials and $Y$ be the number of unsuccessful ones (we count as a run any sequence of one or more identical outcomes). Find

(b) Mean lengths of both runs, i.e. $E(X)$ and $E(Y)$.

(c) The correlation function of $E(XY)$.

(d) The covariance $\operatorname{Cov}(X,Y)$.

I am confused as I know it is a geometric distribution, and get confused as to how to go about doing this :/

Thanks for any help.

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Or I think it may be a Negative binomial distribution ... help please :) – TOmmy Lee Nov 18 '12 at 16:14
There is some non-standard language used here, I think you may have changed some of the words. At one stage $X$ is the number of successful trials. Later, $X$ seems to have something to do with the length of a run. I suspect that you are actually being asked about the number of runs. The definition of $X$ needs to be clarified. – André Nicolas Nov 18 '12 at 17:43
I can see three reasonable ways the question might be construed. (1) It could be about conditional distributions given that the first trial results in success. Then you'd have some number $X\ge1$ of consecutive successes followed by some number $Y\ge1$ of consecutive failures. (2) It may be that we allow $\Pr(X=0)>0$, so that the first trial may result in failure. (3) Maybe one should call whichever of the two outcomes occurs first "success" and the other "failure". Then $X$, $Y$ would necessarily be positive. – Michael Hardy Nov 18 '12 at 18:34
I just rejected an anonymous edit from a "fellow student" which said: X is solely the number of successful trials, in regards to the runs this is just to say that it isn't a binomial distribution or a geometric where it means we stop at the first success. Basically to hint at the type of distribution it is. – Asaf Karagila Nov 18 '12 at 22:14
The following seems to be clear: one should condition on the first trial being a success; part (c) is meaningless; the OP did not show anything about what they tried; this is (homework). – Did Nov 19 '12 at 8:29