Can var(x+y) Be Less Than or Equal to 2(var(x) + var(y))?

[kbilsback5]

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!

 

[noowutah]

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

$2\gamma\leq{}2(\gamma^{2})^{\frac{1}{2}}\leq{}2(\alpha\beta)^{\frac{1}{2}}\leq{}2(\frac{\alpha{}+\beta}{2})\leq{}\alpha{}+\beta$

where $(\alpha\beta)^{\frac{1}{2}}\leq{}\frac{\alpha{}+\beta}{2}$ 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).

 

[kbilsback5]

Awesome! Thanks so much for your help!

 

[noowutah]

You are welcome.

 

[blue_raver22]

I am happy to assist with this proof. Let's start by defining the var(x+y) inequality as follows:

var(x+y) = E[(x+y)^2] - [E(x+y)]^2

= E[x^2 + 2xy + y^2] - [E(x) + E(y)]^2

= E(x^2) + 2E(xy) + E(y^2) - [E(x)]^2 - 2E(x)E(y) - [E(y)]^2

= [E(x^2) - [E(x)]^2] + [E(y^2) - [E(y)]^2] + 2[E(xy) - E(x)E(y)]

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

Now, we can use the definition of covariance to rewrite this as:

var(x+y) = var(x) + var(y) + 2[E(xy) - E(x)E(y)]

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

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

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

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

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

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

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

Now, since the covariance is always less than or equal to the product of the standard deviations of x and y, we can write:

var(x+y) ≤ var(x) + var(y) + 2[σ(x)σ(y)]

= var(x) + var(y) + 2√[var(x)var(y)]

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

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

Therefore, we have proven that var(x+y) ≤ 2(var(x) + var(y)). I hope this helps!