If we know that X is normally distributed, how would we show that cX+d is normally distributed, for c and d real numbers? In particular if X is a standard distribution (i.e. with mean 0 and variance 1)?
We haven't done moment generating functions yet, so I don't think I'm expected to do it that way.
How would the substitution work? I tried putting it into the standard 1/2pi... formula, but wasn't able to see how to rearrange it to give another normal distribution.
Let Y=cX+d.
Thus X=(Y-d)/c.
Note that c is not zero.
Next get the derivative, which is dx/dy= 1/c.
Then place the (Y-d)/c in for X in the exponent of e in the density of X.
Mulitiply that by the ABSOLUTE value of the derivative 1/|c|.
and you will see that the density is now a normal with mean cEX +d
and variance c^2 V(X), after you play with the exponent a bit.
It's just
f(y)=f(x)|dx/dy|, where these are two different f's.
And that's just calc one change of variables, where you do need the absolute value.
We can't have negative densities.
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((y-d)/c-mu)^2= (y-d-cmu)^2/c^2 is the next step.
Then place the c^2 in with sigma^2 and it's over.