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Two random variables, $X$ and $Y$ , have the joint distribution $P(x, y)$, $$ \begin{array}{cc|cc} && x\\ && 0 & 1\\ \hline y & 0 &0.5 &0.2\\ &1 &0.2& 0.1 \end{array}$$

  1. Are $X$ and $Y$ independent? Explain.
  2. Are $(X + Y )$ and $(X − Y )$ independent? Explain.

How to prove that $X+Y$ and $X-Y$ are independent?

Arthur
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Satyaki Chatterjee
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  • The same way you would any other random variables: Either prove that the results of one doesn't affect the probabilities of the other, or give a counterexample. – Arthur Jan 12 '18 at 12:23
  • In case of P(X,Y)=P(X) * P(Y) we have pmf of L.H.S to check. But in this case how will i calculate pmf of P(X+Y,X-Y) – Satyaki Chatterjee Jan 12 '18 at 12:27
  • $P(0,0)$, for instance, means the probability of both $X + Y = 0$ and $X-Y = 0$, so $X = Y = 0$, which means $P(0,0) = 0.5$. Next, $P(1, 0)$ is the probability that $X+Y = 1$ and $X-Y = 0$, which has no valid solutions, so $P(1, 0) = 0$. And so on. – Arthur Jan 12 '18 at 12:38

1 Answers1

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$P(X)=0.2+0.1=0.3$ $P(Y)=0.2+0.1=0.3$ $$P(XY)=0.1 \neq P(X)P(Y) = 0.09 $$

Thus they are not independent. Similarly do the same thing for $X+Y$ and $X-Y$

avz2611
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  • Sir, i having some problem calculating it,will you please do that for me? The problem is i got P(XY) form the table given above..but how will i be able to find P(X+Y,X-Y) form here? – Satyaki Chatterjee Jan 12 '18 at 12:35