Re: How to prove this

"Dirk Van de moortel" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
>
> "Cool Giraffe" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED]
> > Suppose that the density function of X is:
> > f(x) = ( e^-x * x^a ) / a! ,   x > 0 and a in N.
> >
> > Prove that:
> > P( 0 < X < 2a + 2 )   >=   a / ( a + 1 ).
> >
> > I have no idea how to start (using a straight-forward
> > method and integrating f(x) doesn't give anything fun.
> > A hint is to use "a proper inequality" but that's of no
> > use for me either, since i'm hopelessly stuck. Help?
>
> Chebychev inequality:
>          P[ |X-mu| >= k ] <= s^2/k^2    for all k
>
> ==>  1 - P[ |X-mu| < k ] <= s^2/k^2    for all k
> ==>  P[ |X-mu| < k ] >= 1 - s^2/k^2    for all k
> ==>  P[ -X+mu < k < X-mu ] >= 1 - s^2/k^2    for all k

silly!
that should be:
==>  P[ -k < X-mu < k ] >= 1 - s^2/k^2    for all k

>
> Take X with a gamma distribution with
>         alpha = a+1
>         beta = 1
>     then
>         mu = alpha = a+1
>         s^2 = alpha*beta^2 = a+1
> Take k = a+1
>
> ==>  P[ -X+a+1 < a+1 < X-a-1 ] >= 1 - (a+1)/(a+1)^2

so that should be:
==>  P[ -a-1 < X-a-1 < a+1 ] >= 1 - (a+1)/(a+1)^2

> ==>  P[ 0 < X < 2a+2 ] >= 1 - (a+1)/(a+1)^2
> ==>  P[ 0 < X < 2a+2 ] >= 1 - 1/(a+1)
> ==>  P[ 0 < X < 2a+2 ] >= a/(a+1)
>
> hth
>
> Dirk Vdm

sorry :-)
Dirk Vdm