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

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.

• Is it stated that $X$ and $Y$ are independent? That is being assumed. Do you know these properties of variance en.wikipedia.org/wiki/Variance#Properties ? Learn them and use them. – Mark L. Stone Jan 16 '17 at 3:08
• Then whoever (book, teacher) provided the question is deficient and doing students an injustice by "teaching" you to implicitly make assumptions, which in many real-world problems are not true and greatly affect the real-world answer. – Mark L. Stone Jan 16 '17 at 3:12
• Do you know how often I see people make completely unjustified assumptions of independence, in situations in which that has a tremendous effect on the answer? A whole heck of a lot out in the real world. – Mark L. Stone Jan 16 '17 at 3:18
• well thankfully I have the brilliant minds of internet to actually teach me about statistics and not have me assume things :D – camnesia Jan 16 '17 at 3:20
• If X an Y are independent and normal then Var(2X-Y)=4Var(X) +Var(Y) and since Var(X) =Var(Y)=4 the answer is 4x4+4=20. However as Mark Stone has pointed out if X and Y are correlated (in this case negatively correlated the Cov(X,Y) would be negative and so the variance for Z=2X-Y will be less than 20. Now the means add together whether or not X and Y are independent. So E(2X-Y)=2E(X)-E(Y)= 2x3-5. – Michael R. Chernick Jan 16 '17 at 6:20

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}

• @giusti I think it's also prudent to note that the answer is also a normal random variable and explain why the linear combination of two normal random variables is itself a normal random variable. – StatsPlease Jan 18 '17 at 4:56
• Good point. Please feel free to edit and add that to the answer. – giusti Jan 18 '17 at 11:34

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^{X+Y}]\\ &\overset{*}{=}\mathbb{E}[e^{X}]\mathbb{E}[e^{Y}]\\ &=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 $Y$.