Is it true that to change the mean and stdev of a normal distribution (with mean 0 and stdev 1), all one has to do is to multiply the current PDF by the new stdev and add the mean?
I.e. I have N(0,1), will 1 + 2*N(0,1) give me a normal distribution with mean 1 and stdev 2?
Is so, could someone kindly explain to me how this is (from an equation and an intuitive perspective).
Many thanks.
(N.B - Apologies for the basic question - very new to the idea of distributions)
Lets first adress the mean: If $X ~ N(0,1)$ then $X+m$ has basically the same distribution as $X$ but just shifted by $m$. If we have a realization $x_1 , ... ,x_n$ of $X$ then The mean is
$$0 = \frac 1 n \sum_{i=1}^n x_i = \frac 1 n \left(\sum_{i=1}^n x_i+m \right)-m$$
This implies $$\frac 1 n \left(\sum_{i=1}^n x_i+m \right) = m$$
In other words: We find $m$ is the mean of the $x_i+m$ which are realizations of $X+m$.
Similarly we can do the same thing with the variance, we basically 'stretch' the 'range' of values.
Another approach: Let $Y=mX+b$, then you can easily prove that
$E(Y) = mE(X) + b$ and $Var(Y) = m^2 Var(X)$
Note that $SD(Y)=\sqrt{Var(Y)}$.
I also recommend plotting the PDFs of $N(0,1), N(1,1)=N(0,1)+1, N(0,2)=2N(0,1), N(1,2)=2N(0,1)+1$.