Covariance normally distributed variable algebra

Question

I'm generating some variables in R.

One variable: X is normally distributed with mean 0 and variance 1 Second variable: V is normally distributed with mean 0 and variance 1

Third: I'm defining a new variable Z = X + V

I know that the correlation coefficient between X and Z is 1. But, what is the covariance between X and Z? Someone told me that it should be 1 but I don't understand how.

Answer 1

Assuming that $X$ and $V$ are independent we have $Cov(X,Z)=EXZ-EXEZ=E(X(X+V))=EX^{2}-EXEV=1$.

Answer 2

We know co-variance is defined as

$$\mathbb{E}[(\mathbf{X} - \mathbb{E}[\mathbf{X}])(\mathbf{Z} - \mathbb{E}[\mathbf{Z}])]$$

Which in general is equal to

$$\mathbb{E}[\mathbf{XZ}]-\mathbb{E}[\mathbf{X}]\mathbb{E}[\mathbf{Z}]$$

By linearity of expectation we know that $\mathbb{E}[\mathbf{Z}] = 0$

So we are left to calculate $\mathbb{E}[\mathbf{XZ}]$ or $\mathbb{E}[\mathbf{X^2}]+\mathbb{E}[\mathbf{XY}]$. By linearity of expectation we get $\mathbb{E}[\mathbf{X}^2]$.

$$ \mathbb{E}[\mathbf{X}^2] = \int\limits_{\mathbb{R}} \frac{x^2}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}dx$$

Using the Poisson trick we get

$$ \begin{align} \left(\int\limits_{\mathbb{R}} \frac{x^2}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}d\mu\right)^2 &= \frac{1}{2\pi}\int\limits_{\mathbb{R}} \int\limits_{\mathbb{R}}(xy)^2 e^{-\frac{1}{2}(x^2 + y^2)} dxdy\\ &= \frac{1}{\pi}\int\limits_{[0,2\pi]}\int\limits_{\mathbb{R}_{\geq 0}} r^5\sin^2(\theta)\cos^2(\theta) e^{-\frac{1}{2}r^2} drd\theta \\ &=\frac{1}{\pi}\int\limits_{[0,2\pi]}\sin^2(\theta)\cos^2(\theta)\int\limits_{\mathbb{R}_{\geq 0}} r^5 e^{-\frac{1}{2}r^2} drd\theta \\ &=\frac{1}{4\pi}\int\limits_{[0,2\pi]}\sin^2(2\theta)\int\limits_{\mathbb{R}_{\geq 0}} r^5 e^{-\frac{1}{2}r^2} drd\theta \\ &=\frac{1}{\pi}\int\limits_{[0,2\pi]}\sin^2(2\theta)\int\limits_{\mathbb{R}_{\leq 0}} u^2 e^{u} du d\theta \\ &=\frac{1}{\pi}\int\limits_{[0,2\pi]}\sin^2(2\theta)\left( \left[ u^2e^u\right]_{-\infty}^0 - \int\limits_{\mathbb{R}_{\leq 0}} u e^{u} du \right) d\theta \\ &=\frac{1}{\pi}\int\limits_{[0,2\pi]}\sin^2(2\theta)\left( \left[ u^2e^u\right]_{-\infty}^0 - \left[ ue^u\right]_{-\infty}^0 + \int\limits_{\mathbb{R}_{\leq 0}} e^{u} du \right) d\theta \\ &=\frac{1}{\pi}\int\limits_{[0,2\pi]}\sin^2(2\theta)\left( \left[ u^2e^u\right]_{-\infty}^0 - \left[ ue^u\right]_{-\infty}^0 + \left[e^{u} \right]_{-\infty}^0 \right) d\theta \\ &=\frac{1}{\pi}\int\limits_{[0,2\pi]}\sin^2(2\theta)\left( 0 + 0 + 1\right) d\theta \\ &=\frac{1}{\pi}\int\limits_{[0,2\pi]}\frac{1}{2} (1 - \cos(4 \theta))d\theta \\ &=\frac{1}{\pi}\left( \pi -\int\limits_{[0,2\pi]}\frac{1}{2} \cos(4 \theta)d\theta \right)\\ &=\frac{1}{\pi}\left( \pi -[\frac{1}{4}sin(4\theta)]_0^{2\pi}\right)\\ &=1\\ \end{align}$$

Taking the square root we get $1$.

The more simple, general, and clever way to do this is simply by the bilinarity of co-variance. Namely, $$\mathbf{COV}(X, X+Y) = \mathbf{COV}(X, X) + \mathbf{COV}(X, Y) = 1 + 0 = 1$$