Caroline Le Gall wrote:
>
> Hello all,
>
> Is someone can help me ?
> If X is distributed as a gaussian distribution with m as mean and s as
> standard deviation, what is the distribution of 1/X ?
Horrible <grin>
> Particularly, what is the mean and the standard deviation of 1/X ?
Hasn't got any!
*No* continuous variable whose probability is nonzero in a neighborhood
of 0 has a reciprocal variable with mean or standard deviation. By
compactness f_X(x) is bounded above by some b>0 on [-a,a]; f_{1/X}(x)
is then bounded below by b/x^2 outside of [-1/a,1/a] and neither
$\int_{1/a}^\infty x f_{1/X}(x) dx $
nor
$\int_{1/a}^\infty x^2 f_{1/X}(x) dx$
converges.
This may seem counterintuitive if the coefficient of determination is
large - if the mean is many times the standard deviation. In reality, IF
the reciprocal has any concrete interpretation, the model will probably
break down before one gets to the super-extreme tails. How you take
this into account to obtain a useful number is up to you.
One obvious solution: if you integrate
$\int_{\mu-a \sigma}^{\mu+a \sigma} x f_{1/X} dx$ for various values of
$a$, and $|\mu|>>\sigma$, the result will be essentially constant over
a significant range of $a$ before it explodes; take the value on the
"plateau" as the effective mean.
-Robert Dawson
.
.
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