If $X \sim \mathcal N(0,1)$ and $Y \sim \mathcal N(0,1)$ are i.i.d. standard normal distributed how can I find:

  • $W=3X+Y-2$
  • $\mathrm{Cov}(X+Y, X-Y)$

  • $\mathbb{P}(X\lt2Y)$


Not sure:

$W\sim\mathcal N(-2,10)$


$\mathrm{Cov}(X+Y, X-Y)=0$


I guess that $\mathbb{P}(X\lt2Y) = \mathbb{P}(X-2Y\lt0)$, so if $X-2Y\sim \mathcal N(0,5)$ then $$\mathbb{P}(X-2Y\lt0)=\mathbb{P}(Z \leq \frac{0-0}{\sqrt 5})=\Phi(0)=0.5$$


Thanks to Andrè and Michael for helping.

  • $\begingroup$ I imagine for first you mean normal mean $-2$ variance $10$. For the second, you should get $0$, variance of $X$ minus variance of $Y$. $\endgroup$ Jun 23 '16 at 15:50

Q1: If $X$ and $Y$ are independent normal, and $W=aX+bY+c$, then $W=aX+bY+c$ is normal, mean $aE(X)+bE(Y)+c$ and variance $a^2\text{Var}(X)+b^2\text{Var}(Y)$.

Q2: We use the bilinearity of covariance. We have $\text{Cov}(X+Y,X-Y)=\text{Cov}(X,X-Y)+\text{Cov}(Y,X-Y)$.

But $\text{Cov}(X,X-Y)=\text{Cov}(X,X)-\text{Cov}(X,Y)$. This is $\text{Var}(X)-\text{Cov}(X,Y)$. The second term is $0$ by independence.

Similarly, $\text{Cov}(Y,X-Y)=-\text{Var}(Y)$.

So we end up with $\text{Var}(X)-\text{Var}(Y)$, which is $0$.

Another way: We want $E((X+Y)(X-Y))-E(X+Y)E(X-Y)$. The first term is $E(X^2)-E(Y^2)$, which in our case is $0$, and the second term, in our case, is $0$.

This is faster than the bilinearity of covariance approach. I mentioned that one first because the bilinearity is a useful fact. It is likely that you are intended to take the second approach.


In $\Pr(X-2Y<0)=\Pr\left(Z \leq \frac{0-0}{5}\right)$ you need $\sqrt 5$ rather than just $5$ in the denominator. You divide by the standard deviation, not by the variance.

For the first question you need $W\sim\mathcal{N}(-2,10)$. For the normal distribution you specify the expected value and the variance to say which normal distribution it is.

Otherwise ok.


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