# Why is the variance of this random variable one?

I stumbled across the following problem in the paper "STOCHASTIC SPANNING TREE PROBLEM" of Ishii. There is a random variable defined by $$\frac{\sum_j (c_j x_j - \mu_jx_j)}{(\sum_j\sigma_j^2x_j^2)^{1/2}},$$ where each $$c_j$$ is distributed according to a normal distribution with mean $$\mu_i$$ and variance $$\sigma_j^2$$. It is said that this variable is distributed according to the standard normal distribution and I try to proof this for myself.

For the expected value I get zero, just by using its linearity and then the sums in the numerator cancel out. Now to the variance: I try to use the property that $$\text{Var}(X) = \mathbb{E}(X^2) - (\mathbb{E}(X))^2$$ where I already know that the second term is zero. Then I square the fraction and aim to get a nominator that cancels out to one with the denominator. Since the denominator is constant I drag it out of the expected value and get $$\frac{\mathbb{E} \left[ \left( \sum_j (c_j x_j) - \sum_j(\mu_jx_j) \right)^2 \right]}{\sum_j\sigma_j^2x_j^2}$$ which I expand to $$\frac{\mathbb{E} \left[ \left( \sum_j c_j x_j \right)^2 -2 \left( \sum_j c_j x_j \right)\left( \sum_j \mu_j x_j \right) + \left(\sum_j\mu_jx_j \right)^2 \right]}{\sum_j\sigma_j^2x_j^2}$$ where I again use the linearity of the expected value to get $$\frac{\mathbb{E} \left[ \left( \sum_j c_j x_j \right)^2 \right] - \left(\sum_j\mu_jx_j \right)^2}{\sum_j\sigma_j^2x_j^2} = \frac{\mathbb{E} \left[ \left( \sum_j c_j x_j \right)^2 - \left(\sum_j\mu_jx_j \right)^2\right]}{\sum_j\sigma_j^2x_j^2}$$ and now I am stuck and don't know what I miss.

I hope somebody can bring ligth into the dark.

T3 K14

• You need to use independence (or at least pairwise independence) of the $c_j$s. I assume they are independent? (If so, what you want is that variancenif sum = sum of variances when the covariance is 0) – Clement C. Jan 28 at 22:21

Independence of $$c_j$$ is needed.
$$(\sum_jc_jx_j)^2=\sum_jx_j^2c_j^2+\sum_j\sum_{i\ne j}x_ix_jc_ic_j$$
$$(\sum_j\mu_jx_j)^2=\sum_jx_j^2\mu_j^2+\sum_j\sum_{i\ne j}x_ix_j\mu_i\mu_j$$
For $$i\ne j$$, $$E(c_ic_j)=\mu_i\mu_j$$ (independence) so double sum over $$i,j$$ is zero.
$$E(x_j^2(c_j^2-\mu_j^2))=x_j^2(E(c_j^2)-\mu^2)=x_j^2\sigma_j^2$$, so sum over $$j$$ in the numerator is identical to the denominator.