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Let $x$ be a R.V. that is normally distributed with mean 0 (for simplicity we can even assume its a standard normal).

Let $F$ be a monotonic function such that $F(0)=0$

Does it follow that y=F(x) is also normally distributed? Does it even follow that if we graphed the pdf of y, the peak would be at 0.

I would think so...

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Not me... $ $ $ $ – Did May 24 '12 at 20:59
Not necessarily normal. Actually, hardly ever normal. – André Nicolas May 24 '12 at 20:59
Take $F(x)=0$ identically – Student May 24 '12 at 21:00
up vote 1 down vote accepted

By altering $F$, $Y=F(X)$ can have any distribution you like so long as the (or at least a) median of $Y$ is $0$.

So it can have a normal distribution (and will be if $F(x)=kx$ for some non-zero $k$), but can be any other shape.

The peak(s) can also be anywhere. Take for example $F(x) = 4k(\Phi (x))^2 -k$ where $\Phi(x)$ is the cumulative distribution function of a standard normal so $\Phi(0)=\frac12$ and $F(0)=0$. $Y=F(X)$ will have a mode of $-k$.

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How do we know that about the mode? – David May 24 '12 at 22:57
@David: In my example, $Y$ would have a density of $\dfrac{1}{\sqrt{|16k(y+k)|}}$ between $y=-k$ and $y=3k$, so with an arbitrarily high density near $y=-k$. – Henry May 25 '12 at 7:13

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