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Given Problem

Let $X_1, X_2, ..., X_n$ be independent, normally distributed random variables having mean $0$ and variance $\sigma^2$. How should $P(\sqrt{X_1} + \sqrt{X_2} + ... + \sqrt{X_n} \leq x)$ be approximated in terms of $\Phi$?

Thoughts for the Problem

From the moment generating function, we see that if $X$ is a random variable of the normal distribution, then

$$E[X^p] = \left\{ \begin{array}{c c} 0 & \text{if $p$ is odd}\\ \sigma^p (p - 1)!! & \text{if $p$ is even} \end{array} \right. $$

For each $p$ (given the mean $0$ and variance $\sigma^2$), we have

$$p = 1 \rightarrow 0$$ $$p = 2 \rightarrow \sigma^2$$ $$p = 3 \rightarrow 0$$ $$p = 4 \rightarrow 3\sigma^4$$ $$p = 5 \rightarrow 0$$

...and so on. These central moments are found by the following formulas

$$M_X(t) = e^{\mu t + \frac{1}{2} \sigma^2 t^2}$$

where $\mu = 0$, and

$$M_X^{(n)}(0) = E[X^n]$$

which is used to determine the mean of $X^n$.

Let's suppose we want to determine the mean and variance of the random variable $X_i^{\frac{1}{2}}$ for each $i = 1, 2, 3, ..., n$. Then,

$$E\left[X^{\frac{1}{2}} \right] = M_X^{\left(\frac{1}{2} \right)}(0)$$

and

$$Var\left[X^{\frac{1}{2}} \right] = E[X] - \left(E\left[X^{\frac{1}{2}} \right] \right)^2$$

Here is the part, where I am stuck. How does one compute fractional derivative of the moment generating function at $t = 0$? Is there a simple way to work this out? Also, would doing the approximation acquires the same steps as approximating the sum of variables with integer exponents?

I have a lot of thoughts about this problem. My professor gave the problem with variables containing fractional exponents instead of integer exponents.

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4  
How do you define $\sqrt{X}$ when $X<0$? (if $X \sim N(0,\sigma^2)$, it can obviously take negative values) –  Jean-Claude Arbaut Dec 20 '13 at 21:14
    
Oh wow. Didn't see this coming. Thanks for the hint. You really gave it away. –  NasuSama Dec 20 '13 at 22:02
    
Now, I need to define the distribution of $\sqrt{X}$ and work out the mean and variance, treating the "new" distribution with $\sqrt{X}$. This should be extremely simple –  NasuSama Dec 20 '13 at 22:03
2  
But there is still this problem. Just in case you would like to take a complex square root, don't expect to be able to satisfy the inequality $\leq x$. You know, inequalities in $\Bbb C$... Obviously there is something wrong: either take $\sqrt{|X|}$, or take another distribution for $X$. –  Jean-Claude Arbaut Dec 20 '13 at 22:07

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