Valid distribution If $N(0,\sigma^2)$ is the Gaussian distribution with mean $0$ and variance $\sigma^2$, is $pN(0,\sigma^2)$ a valid distribution ? $p$ is a constant and $0\le p\le 1$.
 A: To answer this, I have to guess what you mean by the question.  I have two guesses.  Assuming one of the two, the answer is "no", and assuming the other, it's "yes".
The normal distribution assigns a probability to each interval from a number $a$ to a number $b$.  Let's say $X$ is a random variable whose probability distributino is $N(0,\sigma^2)$.  Then $\Pr(a<X<b)$ is determined by that distribution and which numbers $a$ and $b$ are.
If you mean by $pN(0,\sigma^2)$ a measure that just multiplies $\Pr(a<X<b)$ by $p$ and says that's the "probability" of some random variable being between $a$ and $b$, then the answer is "no", because it would say $\Pr(-\infty<X<\infty)=p$, whereas any probability distribution of a real-valued random variable $X$ would say $\Pr(-\infty<X<\infty)=1$.
On the other hand, I wonder if by $pN(0,\sigma^2)$ you might mean the distribution of $pX$, where $X\sim N(0,\sigma^2)$.  In that case, one would say $pX\sim N(0,p^2\sigma^2)$.  And that is a valid probability distribution.
Of course, it's possible that both of my guesses are wrong.
Later note: From comments it seems my first guess was right.  So this is not a probability distribution.  The density of the $N(0,\sigma^2)$ distribution is
$$f(x) = \frac{1}{\sqrt{2\pi\sigma^2\  {}}} \exp\left(\frac{-x^2}{2\sigma^2}\right)  = \frac{1}{\sqrt{2\pi}\  \sigma} \exp\left(\frac{-x^2}{2\sigma^2}\right)$$
And as with all probability densities on the real line, we have
$$\int_{-\infty}^\infty f(x)\,dx=1.$$
But
$$\int_{-\infty}^\infty pf(x)\,dx \ne 1\text{ unless }p=1$$
so it's not a probability density unless $p=1$.  It is, however the density of an absolutely continuous measure.
