A deck of 52 cards is shuffled and a bridge hand of 13 cards is dealt out. Let X and Y denote, respectively, the number of aces and the number of spades in the hand.

(a) Show that X and Y are uncorrelated.

(b) Are they independent?

I know that if ${\rm Cov}(X,Y)=0$ then $X$ and $Y$ are uncorrelated. I also know that $X$ and $Y$ are independent since the probability of choosing aces does not affect the probability of drawing spades. I tried using ${\rm Cov}(X,Y)=E[XY]-E[X]E[Y]$ but got stuck on trying to find the expected value of $X$ and $Y$. Is this a valid approach to the problem or is there something I am missing?

  • $\begingroup$ Perhaps you could try iterated expectations for $E(XY)$. If you think they are independent, then what would the covariance be in part a? To explore the question of whether or not they are independent events, perhaps you could try to calculate $P( X | Y)$ for various values of $Y$ to see if the conditional distribution of $X$ depends on the value of $Y$. $\endgroup$ – jsk May 7 '14 at 2:00
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    $\begingroup$ > I also know that $X$ and $Y$ are independent... You mean that $P\{X=4, Y=13\}$, the probability that the hand contains all $4$ aces and also has all $13$ spades, equals $P\{X=4\}P\{Y=13\}$, the probability of $4$ aces times the probability of all $13$ spades? Since $P\{X=4, Y=13\}= 0 = P\{X=4\}P\{Y=13\}$, which of $P\{X=4\}$ and $P\{Y=13\}$ would you say is $0$? In any case, if you know that $X$ and $Y$ are independent, there is no need to check covariance: independent random variables are uncorrelated random variables (subject to existence of various expectations, of course). $\endgroup$ – Dilip Sarwate May 7 '14 at 3:15
  • $\begingroup$ It might be easier to follow the thrust of @Dilip's comment by noticing that it is based on a rigorous definition of independence of random variables--which requires that all relevant probabilities multiply in the way shown in his example--whereas "does not affect the probability" is neither a definition nor, in fact, is it terribly useful (because it is merely a non-operational intuition used to motivate the definition and one's intuition can easily be confused by questions of probability). $\endgroup$ – whuber May 7 '14 at 15:21
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    $\begingroup$ @whuber Thanks for providing supporting material for my argument. It is unfortunate that for the experiment of drawing one card at random from the deck, the events "card is a spade" and "card is an ace" are indeed independent events of probabilities $\frac 14$ and $\frac{1}{13}$ respectively since their intersection "card is ace of spades" has probability $\frac{1}{52}=\frac 14\times\frac{1}{13}$. However, this does not extend to the experiment of drawing $13$ cards at random without replacement and the independence of the random variables $X$ and $Y$. $\endgroup$ – Dilip Sarwate May 7 '14 at 19:14
  • $\begingroup$ Is $E[X]=4/52$ and $E[Y]=13/52$? $\endgroup$ – Oscar Flores May 7 '14 at 19:59

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