Distribution of 2X - Y when X and Y are known

Question

I have been given a question:

 If X ~ N (3, 2²) and Y ~ N(1, 2²) what is the distribution of 2X-Y ?

The answer is given as N(5,20)

I assume the value for 5 is = 2x-y = (3 x 2) - 1 but where does the 20 come from ? I am very new to statistics and any help would be appreciated.

Answer 1

For two variables $X$ and $Y$, the variance of $aX-bY$ is:

$$Var(aX-bY)=a^2Var(X)+b^2Var(Y)-2ab·Cov(X,Y)$$

Where $Cov(X,Y)$ is the covariance between $X$ and $Y$. (source)

Usually the problem statement should say whether the variables $X$ and $Y$ are dependent or independent. Since this one apparently doesn't, I'm taking a guess and claiming independence—only because that matches the expected problem answer: first, the linear combination of two independent Normal random variables is a Normal random variable; Second, the covariance between two independent variables is zero.

So:

$$\begin{align}Var(2X-Y)&=2^2Var(X)+1^2Var(Y)-2ab·Cov(X,Y)\\ &=4 · Var(X)+Var(Y)-0\\ &=4·4+4\\ &=20\end{align}$$

Answer 2

There are a few important results that are required here. Based on the provided answer, I think an assumption of the question was that $X$ and $Y$ are independent random variables.

Let's try and start by proving an important result

$$X\sim N(\mu,\sigma^{2})\Leftrightarrow aX\sim N(a\mu,a^{2}\sigma^{2})$$

We can do this many ways, but let's use the moment generating function of the normal distribution. Let

$$M_{X}(t)=\mathbb{E}[e^{tX}]=e^{\mu t+\tfrac{1}{2}\sigma^{2}t^{2}}$$ and $$X\sim N(\mu,\sigma^{2})$$

So,

$$M_{aX}(t)=\mathbb{E}[e^{atX}]=e^{a\mu t+\tfrac{1}{2}a^{2}\sigma^{2}t^{2}}$$ and $$aX\sim N(a\mu,a^{2}\sigma^{2})$$

Another important result is the sum of two independent normal random variables.

If $$X\sim N(\mu_{x},\sigma^{2}_{x})$$ and $$Y\sim N(\mu_{y},\sigma^{2}_{y})$$ then $$X+Y\sim N(\mu_{x}+\mu_{y},\sigma_{x}^{2}+\sigma_{y}^{2})$$

Once again we can prove this using the moment generating function:

$$\begin{align} M_{X+Y}(t)&=\mathbb{E}[e^{t(X+Y)}]\\ &\overset{*}{=}\mathbb{E}[e^{tX}]\mathbb{E}[e^{tY}]\\ &=e^{\mu_{x} t+\tfrac{1}{2}\sigma_{x}^{2}t^{2}}e^{\mu_{y} t+\tfrac{1}{2}\sigma_{y}^{2}t^{2}}\\ &=e^{(\mu_{x}+\mu_{y})t+\tfrac{1}{2}(\sigma_{x}^{2}+\sigma_{y}^{2})t^{2}} \end{align}$$

So $$X+Y\sim N(\mu_{x}+\mu_{y},\sigma_{x}^{2}+\sigma_{y}^{2})$$

$^{*}$Due to independence between $X$ and $Y$.