Stochastic ordering I'm a bit stuck with the following.
Assume $X$ ~$N(0,1)$ and $Y$~$N(1,1)$.
Is $Y$ stochastic bigger than $X$?
And the same question goes for $X$~$N(0,1)$, $Y$~$N(0,4)$..
I simply don't have a clue how to approach it. Can one please guide me?
Thanks!
 A: According to Wikipedia:

A real random variable $X$ is smaller than a random variable $Y$ in the "usual stochastic order" if:
$$ \forall \ r \in \mathbb{R}, \ \mathbb{P}(X>r) \leq \mathbb{P}(Y>r) $$

This is equivalent to saying:
$$ \forall \ r \in \mathbb{R}, \ \mathbb{P}(X\leq r) \geq \mathbb{P}(Y\leq r) $$
Now, let $X \sim \mathcal{N}(0,1)$ and $Y \sim \mathcal{N}(1,1)$. We have:
$$ \begin{align}
\mathbb{P}(Y \leq r) & = \int_{-\infty}^r\frac{1}{\sqrt{2\pi}}e^{-\frac{(u-1)^2}{2}}du 
\end{align} $$
Doing the change of variable $u = t+1$:
$$ \begin{align}
\mathbb{P}(Y \leq r) & = \int_{-\infty}^{r-1}\frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}du 
\\[12pt]
& = \mathbb{P}(X \leq r-1)
\\[12pt]
& \leq \mathbb{P}(X \leq r)
\end{align} $$
Because $\mathbb{P}(Y \leq r) \leq \mathbb{P}(X \leq r)$, $X$ is stochastically-smaller than $Y$ $-$ this is rather intuitive if you consider the behaviour of normal random variables and the fact that $\mathbb{E}[Y] \geq \mathbb{E}[X]$.
Now, let $X \sim \mathcal{N}(0,1)$ and $Y \sim \mathcal{N}(0,4)$ $-$ I am assuming the 2nd argument of $\mathcal{N}(\cdot,\cdot)$ is the variance. We have:
$$ \begin{align}
\mathbb{P}(Y \leq r) & = \int_{-\infty}^r\frac{1}{\sqrt{2\pi 2^2}}e^{-\frac{1}{2}\left(\frac{u}{2}\right)^2}du 
\end{align} $$
Doing the change of variable $u = 2t$:
$$ \begin{align}
\mathbb{P}(Y \leq r) & = \frac{1}{2}\int_{-\infty}^{\frac{r}{2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}\frac{du}{2} 
\\[12pt]
& = \frac{1}{4}\mathbb{P}\left(X \leq \frac{r}{2}\right)
\end{align} $$
As pointed out by @Dilip Sarwate and @Mark L. Stone $-$ big mistake on my side... $-$ in this case neither is stochastically-bigger in the sense above $-$ indeed whether $\mathbb{P}(Y \leq r)$ or $\mathbb{P}(X \leq r)$ is bigger will depend on the sign of $r$. Sorry for the mistake! 
Generally speaking, if we let $X \sim \mathcal{N}(0,1)$ and $Y \sim \mathcal{N}(\mu,\sigma^2)$, we have $-$ this can be shown through a change of variables, like above:
$$ \mathbb{P}(Y \leq r) = \frac{1}{\sigma^2}\mathbb{P}\left(X \leq \frac{r-\mu}{\sigma}\right)$$
