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Let $X_1,\dots,X_n$ be independent random variables, each normally distributed as $X_k\sim N(m_k;\sigma^2_k)$. Let $S_n = \sum_{k=1}^n X_k - m_k$ and $T_n = \frac{S_n}{\sqrt{\operatorname{Var}(S_n)}}$.

We wish to show that $\lim_{n\to\infty} T_n \sim N(0;1)$.

I've laboriously proven that $T_n \sim N(m; \sigma^2)$ where $m=\sum_k m_k$ and $\sigma^2=\sum_k \sigma^2_k$. So now I'm wondering if I'm supposed to assume that the $\lim_{n\to\infty}m=0$ and similarly $\sigma^2\to 1$, which I don't really see a strong justification for.

Is this what I'm supposed to do? What's the justification for it?

(Note this is exercise 14.31.7 in Apostol's Calculus II. Looking online it seems like some other places don't define CLT as it approaches standard normal, just normal distribution of some mean and variance. So this could be an unusual definition he's using.)

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Your laborious proof that $T_n\sim N(m;\sigma^2)$ must be wrong unless $m= 0$ and $\sigma^2 = 1$. Just by the definition of $T_n$ you always have $$\sqrt{Var[S_n]} E[T_n] = E[S_n] = \sum_{k=1}^n (E[X_k] - m_k) = 0$$ and $$Var[T_n] = \frac{Var[S_n]}{Var[S_n]} = 1$$ But a $N(m;\sigma^2)$-distributed random variable has mean $m$ and variance $\sigma^2$. – Sam May 26 '12 at 13:33
Did you maybe calculate the distribution of $X_1 + \dots + X_n$ instead of the one for $T_n$? $X_1 + \dots + X_n \sim N(m;\sigma^2)$ should be right. In this case you would only need to show that if $Y \sim N(m;\sigma^2)$, then $Y-m \sim N(0;\sigma^2)$ and $(Y-m)/\sigma \sim N(0;1)$. – Sam May 26 '12 at 13:35
@SamL: That is indeed what I did, thanks! – Xodarap May 26 '12 at 14:06
up vote 1 down vote accepted

You don't need these assumptions. In fact, $E[S_n]=0$ and $\operatorname{Var}(T_n)=1$. Furthermore, we know that a sum of two independent random variable $Y_1$ and $Y_2$ normally distributed, respectively $Y_1\sim \mathcal N(m_1,\sigma_1^2)$, $Y_2\sim \mathcal N(m_2,\sigma_2^2)$ then $Y_1+Y_2\sim\mathcal N(m_1+m_2,\sigma_1^2+\sigma^2_2)$. So by induction we deduce that $T_n\sim\mathcal N(0,1)$ for each $n$. Convergence in law follows now from the definition.

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Each $X_k-m_k$ is normal and has mean 0 and $S_n$ is $N(0, Var(S_n))$ so $\frac{S_n}{\sqrt{Var(S_n)}}$ is $N(0,1)$.

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