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Thread: Check Answer: Show that two Bernoulli random variables are independent

  1. #1
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    Check Answer: Show that two Bernoulli random variables are independent

    I've found a good answer to a particular question I was given, though I'd like a second opinion on whether or not my answer is solid (I might be "begging the question" in this).

    Show that two Bernoulli random variables $\displaystyle X$ and $\displaystyle Y$ are independent if and only if $\displaystyle P(X=1,Y=1)=P(X=1)P(Y=1)$.

    Here's my answer.

    Two Bernoulli random variables are independent if and only if they are uncorrelated, and thus have a covariance of zero ($\displaystyle Cov(X,Y)=0$).
    Let $\displaystyle p_x$ be the pmf of $\displaystyle X$, and let $\displaystyle p_y$ be the pmf of $\displaystyle Y$.
    If $\displaystyle X$ and $\displaystyle Y$ are independent then, by definition,
    $\displaystyle Cov(X,Y)=p_{xy}-p_xp_y=P(X=1,Y=1)-P(X=1)P(Y=1)=0$,
    as $\displaystyle P(X=1,Y=1)=P(X=1)P(Y=1)$.
    If on the other hand we have that $\displaystyle Cov(X,Y)=0$, then
    $\displaystyle p_{xy}-p_xp_y=0\Rightarrow p_{xy}=p_xp_y$.
    Therefore, $\displaystyle X$ and $\displaystyle Y$ are independent.

    Are there any mess-ups I might have made somewhere? I got this answer from this pdf file.
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  2. #2
    MHF Contributor matheagle's Avatar
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    you would just have to show that the other 3 hold as well.

    You have $\displaystyle P(X=1,Y=1)=P(X=1)P(Y=1)$

    and you will need $\displaystyle P(X=1,Y=0)=P(X=1)P(Y=0)$
    $\displaystyle P(X=0,Y=1)=P(X=0)P(Y=1)$
    $\displaystyle P(X=0,Y=0)=P(X=0)P(Y=0)$

    and they must sum to one via the total and the marginals, so that should help.
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  3. #3
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    Quote Originally Posted by matheagle View Post
    you would just have to show that the other 3 hold as well.

    You have $\displaystyle P(X=1,Y=1)=P(X=1)P(Y=1)$

    and you will need $\displaystyle P(X=1,Y=0)=P(X=1)P(Y=0)$
    $\displaystyle P(X=0,Y=1)=P(X=0)P(Y=1)$
    $\displaystyle P(X=0,Y=0)=P(X=0)P(Y=0)$

    and they must sum to one via the total and the marginals, so that should help.
    Or, as an alternative, one could do it for $\displaystyle P(X=i,Y=j)=P(X=i)P(Y=j)$ where $\displaystyle i,j=0,1$ (this would be a bit faster, yet still means the same thing). Note that I listed the question word-for-word, and the question doesn't say anything about marginal pmfs or totals. Though I assume one would want to show that the sum of the marginals equals 1 for completeness, I don't know if such would be necessary for the purpose of the question.

    See, I got my answer from this link here, and the proof it gave looks very solid. Is this source missing some necessary parts?
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  4. #4
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    EDIT: This had a doublepost, oops.

    But as long as it's here, I figured out what you mean by that "summing" part. Here's what I got from PhysicsForum.

    $\displaystyle Cov(X,Y)= E(XY)-E(X)E(Y)$
    $\displaystyle = \left(0 \cdot 0 \cdot p(0,0) + 0 \cdot 1 \cdot p(0,1) + 1 \cdot 0 \cdot p(1,0) + 1 \cdot 1 \cdot p(1,1)\right) - \left(0 \cdot p_x(0) + 1 \cdot p_x(1)\right) \left(0 \cdot p_y(0) + 1 \cdot p_y(1)\right)$
    $\displaystyle = p(1,1) - p_x(1)p_y(1)$

    Using this, I'd go into the stuff I put in earlier and be able to complete the proof.

    If there's anything I've missed, let me know.
    Last edited by Runty; Nov 11th 2010 at 01:05 PM.
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