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From: TheGist on 18 Apr 2010 21:52
Suppose X has a pdf that is symmetric about b; i.e.,
f(b+x)=f(b-x) for all -oo < x < oo.
Show that Y=X-b is symmetrically distributed about 0.

My solution so far:
So for the distribution of X which has pdf f(x)
g^-1(y)=y+b
The Jacobian J=dx/dy=1
f(y)=f_x(g^-1(y))*|J|=f(y)

Is that right?
Ok, so now what? How do I complete showing this?
From: Ray Vickson on 19 Apr 2010 04:41
On Apr 18, 6:52 pm, TheGist <theg...(a)nospam.net> wrote:
> Suppose X has a pdf that is symmetric about b; i.e.,
> f(b+x)=f(b-x) for all -oo < x < oo.
> Show that Y=X-b is symmetrically distributed about 0.
>
> My solution so far:
> So for the distribution of X which has pdf f(x)
> g^-1(y)=y+b
> The Jacobian J=dx/dy=1
> f(y)=f_x(g^-1(y))*|J|=f(y)

You are making this way too complicated: there is no need for
Jacobians, or whatever. However, you do need to be careful about
notation, and not use the same "f" to mean two different things in the
same sentence or paragraph. We have F_Y(y)= P{Y <= y} = P{X-b <= y} =
P{X <=y+b} = F_X(y+b), so f_Y(y) = dF_Y(y)/dy = (d/dy)F_X(y+b) = f_X(y
+b). Thus, f_Y(-y) = f_X(b-y) = f_X(b+y) = f_Y(y).

R.G. Vickson


>
> Is that right?
> Ok, so now what? How do I complete showing this?

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