# How to proof variance of the sum $x+y$ is the sum of variance $\sigma_x^2$ of $x$ and the variance $\sigma_y^2$ of $y$?

Show that the variance of the sum $$x + y$$ of the random variable $$x$$ and the random variable $$y$$ is the sum of the variance $$\sigma_x^2$$ of $$x$$ and the variance $$\sigma_y^2$$ of $$y$$?

My attempt:

Is my first time doing statistics and there are alot of terms i am quite unfamiliar with.

Random variable: Value of a variable result from a random experiment: e.g. (the score when a die is rolled once)

Variance($$\sigma^2$$) : Average of squared deviations from the mean or square of standard deviation?

I know variance of sum $$x + y$$ means $$var(x+y) =$$E(x+y)^2 -$$(E(x+y))^2$$ But what does sum of the variance $$\sigma_x^2$$ of $$x$$ and the variance $$\sigma_y^2$$ of $$y$$ means?

Edited: My 2nd attempt: Var(x + y ) = E$$(x+y)^2$$ - $$(E(x+y)^2)$$ = E($$x^2$$ + 2xy +$$y^2$$) - [($$Ex)^2$$ + 2ExEy + ($$Ey)^2$$ ] = [E$$x^2$$ - (E$$x)^2$$ ] + [E$$y^2$$ - (E$$y)^2$$ ] + 2[Exy - ExEy] = Var x + Var y + 2cov (x,y)

Now i assume(not sure is it correct) sum of the variance $$\sigma_x^2$$ of $$x$$ and the variance $$\sigma_y^2$$ of $$y$$

= var ($$\sigma_x^2$$ ) + var($$\sigma_y^2$$) = E($$\sigma_x^2$$ ) -[E($$\sigma_x)]^2$$ + E($$\sigma_y^2$$ ) -[E($$\sigma_y)]^2$$

Since var(x) = $$\sigma_x^2$$,

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

I did not further solve it because i realise is wrong as the above does not have an expression that has an xy expression. So where have i done wrong?

• It means $E((X+Y)^2-(E(X+Y))^2$ instead. – Michael Hoppe Jan 30 at 13:49
• @MichaelHoppe sorry, just now i made a mistake about variance of sum x + y. Anyways what you said i believe is referring to var(x+y) ? How about the case for the sum of variance σx^2 of x and the variance σy^2 of y? What is the expression? – john Jan 30 at 13:56
• – Did Jan 30 at 17:16

The variance of $$X$$ is $$E\left[\big(X - E[X]\big)^2\right]$$ so here we are looking at
$$E\left[\big((X+Y) - E[X+Y]\big)^2\right] \\= E\left[\big((X-E[X])+(Y-E[Y])\big)^2\right] \\ = E\left[\big(X - E[X]\big)^2\right] +E\left[\big(Y - E[Y]\big)^2\right] +2E\left[\big(X - E[X]\big)\big(Y - E[Y]\big)\right]$$
You actually want the result to be the two left terms. The right-hand term cancels out if and only if the covariance is $$0$$, which will happen when $$X$$ and $$Y$$ are independent or in a few other special cases