Find the distribution of $Y-X$.

Question

Let $X \sim N(0,1)$ and $Y \sim N(0,2)$ respectively, independently of each other. Find the distribution of $Y-X$.

How do I proceed? As I am a beginner I don't know how to approach. Please help me in this regard.

Thank you very much.

Answer 1

Hint: Let $Z=Y-X$. Then the PDF of $Z$ is $$f_Z(z)=\frac{dF_Z(z)}{dz}$$ where $$F_Z(z)=\text{Pr}\left[Z=Y-X\leq z\right]=E_X\left\{\text{Pr}\left[Y\leq z + x\big|X=x\right]\right\}$$

Answer 2

Since you say you have a pure math background, I will give you more than you need to see where some of this comes from. The hard way:

The probability density function (pdf) of the sum (or difference) of two independent random variables is the convolution of their pdf's, in the following sense:

If $p(x)$ is the pdf of $X$, $q(y)$ is the pdf of $Y$, $r(z)$ is the pdf of $Z=Y+X$ and $s(w)$ is the pdf of $W=Y-X$, then

$$r(z) = \int_{-\infty}^{\infty} p(x)q(z-x)\ dx$$ $$s(w) = \int_{-\infty}^{\infty} p(x)q(w+x)\ dx$$

The convolution of two Gaussian functions is a Gaussian.

The sum of two independent normally distributed random variables is a normally distributed random variable.

The mean $E[Y-X] = E[Y] - E[X]$

The variance $E[(X-Y)^2] - E[X-Y]^2 = (E[X^2] - E[X^2]) + (E[Y^2] - E[Y]^2)$ is the sum of the variances.

If you haven't seen this before, you can multiply them out, and note that since they are independent, their co-variance is $0$, i.e. $E[XY] - E[X]E[Y] = 0$

$Y-X$ is normally distributed with mean $0$ and variance $3.$