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The var(x+y) Inequality Proof

  1. Apr 8, 2013 #1
    Hi, I was hoping that someone might be able to please help me with this proof.

    Prove that var(x+y) ≤ 2(var(x) + var(y)).

    So far I have:

    var(x+y) = var(x) + var(y) + 2cov(x,y)

    where the cov(x,y) = E(xy) - E(x)E(y), but I'm not really sure to go from there.
    Any insight would be very helpful!

    Thanks!
     
  2. jcsd
  3. Apr 8, 2013 #2
    Let [itex]\alpha=var(x),\beta=var(y),\gamma=cov(x,y)[/itex]. According to the Cauchy-Schwarz inequality (see http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality for the proof), [itex]\gamma^{2}\leq{}\alpha\beta[/itex]. We want to show [itex]\alpha{}+\beta{}+2\gamma\leq{}2(\alpha{}+\beta)[/itex], which follows directly from

    [itex]2\gamma\leq{}2(\gamma^{2})^{\frac{1}{2}}\leq{}2(\alpha\beta)^{\frac{1}{2}}\leq{}2(\frac{\alpha{}+\beta}{2})\leq{}\alpha{}+\beta[/itex]

    where [itex](\alpha\beta)^{\frac{1}{2}}\leq{}\frac{\alpha{}+\beta}{2}[/itex] follows from the well-known fact that the geometric mean is always smaller than the arithmetic mean (see http://www.cut-the-knot.org/pythagoras/corollary.shtml for proof).
     
  4. Apr 8, 2013 #3
    Awesome! Thanks so much for your help!
     
  5. Apr 8, 2013 #4
    You are welcome.
     
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