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Suppose that we have $2n$ iid random variables $X_1,…,X_n,Y_1,…,Y_n$ where $n$ is a large number. I want to find $P((k∑_iX_iY_i+(∑_iX_i)(∑_jY_j))<c)$ for any integer c.

Since $n$ is a large number and all the random variables are $iid$, using central limit theorem, we can say that $k∑_iX_iY_i$, $(∑_iX_i)$ and $(∑_jY_j)$ are approximately normal random variables and $(∑_iX_i)$$(∑_jY_j)$ is the product of two normal random variables which would have Normal Product Distribution.

So $k∑_iX_iY_i+(∑_iX_i)(∑_jY_j)$ is the sum of one normal and one normal product random variable which are dependent.

Now the question is how can we find $P((k∑_iX_iY_i+(∑_iX_i)(∑_jY_j)) \le c)$ for any integer c?

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2 Answers 2

$$Z = \sum_{i=1}^n \sum_{j=1}^n X_i Y_j = \left(\sum_{i=1}^n X_i\right)\left(\sum_{j=1}^n Y_j\right)$$ If $n$ is large, $S_X = \sum_i X_i$ and $S_Y = \sum_j Y_j$ are approximately normal. They have means $n\mu$ and standard deviations $\sqrt{n} \sigma$ where each $X_i$ and $Y_j$ have mean $\mu$ and standard deviation $\sigma$. Of course they are independent. Thus $E[Z] = E[S_X] E[S_Y] = n^2 \mu^2$ and $E[Z^2] = E[S_X^2] E[S_Y^2] = (n^2 \mu^2 + n \sigma^2)^2$, so the variance of $Z$ is $\text{Var}(Z) = E[Z^2] - E[Z]^2 = n^2 \sigma^4 + 2 n^3 \sigma^2 \mu^2$.

The moment generating function of the product of independent normal random variables with means $n\mu$ and standard deviations $n \sqrt{\sigma}$ has, according to Maple, moment generating function $$ M_Z(t) = E[e^{tZ}] = \frac{1}{\sqrt{1 - n^2 \sigma^4 t^2}} \exp\left(\frac{n^2 \mu^2 t}{1 - n \sigma^2 t}\right)$$ for $t < 1/(n \sigma^2)$.

EDIT: If $\mu \ne 0$, it would be better to separate out the effect of the mean. So let $X_i = \mu + \sigma U_i$ and $Y_i = \mu + \sigma V_i$, where $U_i$ and $V_i$ have mean $0$ and standard deviation $1$. Then $$Z = n^2 \mu^2 + n \mu \sigma \sum_{i=1}^n (U_i + V_i) + \sigma^2 \sum_{i=1}^n \sum_{j=1}^n U_i V_j$$ Now $n \mu \sigma \sum_{i=1}^n (U_i + V_i)$ is approximately normal with mean $0$ and standard deviation $\sqrt{2} n^{3/2} \mu \sigma$, while $\sigma^2 \sum_{i=1}^n \sum_{j=1}^n U_i V_j$ has mean $0$ and standard deviation $n \sigma^2$. For large $n$ this term is negligible compared to the $n^{3/2}$ term. So a good approximation to the distribution of $Z$ is normal with mean $n^2 \mu^2$ and standard deviation $\sqrt{2} n^{3/2} \mu \sigma$.

You asked about $ (k−1) \sum_i X_i Y_i+ Z$: call this $(k-1) T + Z$. If we separate out the effect of the mean, $$T = n \mu^2 + \mu \sigma \sum_{i=1}^n (U_i + V_i) + \sigma^2\sum_{i=1}^n U_i V_i$$ where $\mu \sigma \sum_{i=1}^n (U_i + V_i)$ has mean $0$ and standard deviation $\sqrt{2n} \mu \sigma$ and $\sigma^2 \sum_{i=1}^n U_i V_i$ has mean $0$ and standard deviation $\sqrt{n} \sigma^2$. Again, these terms are negligible compared to the $n^2$ and $n^{3/2}$ terms.

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what if we have ∑i∑j f(i,j) XiYj, where f(i,j)=k if i=j (k is an integer) otherwise f(i,j)=1? –  May Nov 16 '12 at 2:48
    
That is $(k-1) \sum_i X_i Y_i + (\sum_i X_i)(\sum_j Y_j)$. What do you want to know about it? –  Robert Israel Nov 16 '12 at 6:06
    
I want to know what its distribution probability function is. –  May Nov 16 '12 at 23:14
    
so using central limit theorem we can say that $(k−1)∑_iX_iY_i$ is a normal random variable and $(∑_iX_i)(∑_jY_j)$ is the product of two normal random variables Normal Product Distribution ? so it is the sum of one normal and one normal product random variable? –  May Nov 16 '12 at 23:14
    
is there any other way to describe this random variable simpler? –  May Nov 16 '12 at 23:19

@May: if they are independent (as in the first i of iid) then $\sum_i \sum_j X_i Y_j =\sum_i X_i \times \sum_j Y_j$.

For example if they are normally distributed then you have the product of two normal random variables, which is in general not normally distributed and this extends to most other distributions, especially if the sums are over a large number of cases, though if the means are non zero and the standard deviations small compared with the means, this may be difficult to spot.

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