#### JaguarXJS

I am not seeing how to handle this problem. Any help would be appreciated.
Thanks, Jomo

#### Lev888

My guess (it's been a while): since E(x) = 0.5, can we just substitute 0.5 for x?

#### romsek

MHF Helper
I don't have a full solution yet but here's an outline.

You can find the distribution of $X^3$ easily enough and find it's mean and variance.

Then in the limit as $n \to \infty$ both numerator and denominator will tend to normal distributions
with appropriate parameters.

Then you have a ratio distribution of two normals and this I believe is Cauchy. You can find the detals
at the wiki "ratio distribution"

That should give you a Cauchy distribution parameterized by $n$ and you can determine if and to what it converges.

That will be convergence in distribution.

#### romsek

MHF Helper
I don't have a full solution yet but here's an outline.

You can find the distribution of $X^3$ easily enough and find it's mean and variance.

Then in the limit as $n \to \infty$ both numerator and denominator will tend to normal distributions
with appropriate parameters.

Then you have a ratio distribution of two normals and this I believe is Cauchy. You can find the detals
at the wiki "ratio distribution"

That should give you a Cauchy distribution parameterized by $n$ and you can determine if and to what it converges.

That will be convergence in distribution.
So I looked at this in further detail. I ended up with the ratio distribution of two normal distributions with parameters

$Z=\dfrac{X}{Y}$

$X \sim N\left(\dfrac n 4,~\dfrac{9n}{112}\right)$
$Y \sim N\left(\dfrac n 2,~\dfrac{n}{12}\right)$

where the second parameter to the normal distribution is the variance.

There is an enormous formula on the ratio distribution wiki page that I reproduced in Mathematica.
Attached below is a snapshot of the sheet.

JaguarXJS