Thread: Transformation of a normal distribution

1. Transformation of a normal distribution

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)?

2. Originally Posted by TriAngle
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)?
An approach based on the moment generating function of Y = cX + d is probably the simplest way to go.

3. Originally Posted by TriAngle
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)?

The moment generating function techinque is fine.
BUT this is basically calc one.
Just substitute the new variable into the original density and multiply
by the derivative of your substitution,
f1(x)=f2(y)|dy/dx|

4. 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.

5. 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.