How to show that a random variable is NOT normally distributed?

Question

Usually we show that a random variable is normally distributed, but how to show a random variable is NOT normally distributed? For example, this game's strategy is to bet $\$1$ in the $(n+1)^{th}$ game if the total winning up to the $n^{th}$ game is $\le 0$, and we stop when the total winning is $>0$. How to show that $W_i = \sum_{i=1}^n A_i X_i$ does not have a normal distribution?

Suppose $X_1, X_2, \dots$ are independent random variables each with a $N(0, 1)$ distribution. Let $\mathcal F_n$ denote the information in $X_1, \dots, X_n$, and let $A_1, A_2, \dots$ be a sequence of random variables such that for each $j$, $A_j$ is $F_{j-1}$ measurable. Let $W_0 = 0$ and for $n > 0$.

Suppose that

$$A_n = \begin{cases} 1, \text{ if } (n = 1) \text{ or } (A_{n-1} = 1 \text{ and } W_{n-1} \le 0)\\ 0, \text{ otherwise} \end{cases} $$

How to show that for all $n>1$, $W_n = \sum_{i=1}^n A_i X_i$ does not have a normal distribution? What are its expected value $\mathbb E[W_n]$ and $P(W_n>0)$?

Answer 1

Step 1. Let

$$ S_n = \sum_{i=1}^{n} X_i \qquad\text{and}\qquad T = \inf\{n \geq 0 : S_n > 0\}. $$

Then by induction, we can show:

1. $\{A_n = 1 \} = \{ T \geq n \}$, and
2. $W_n = S_{T \wedge n}$, where $a \wedge b = \min\{a, b\}$.

Indeed, it is clear that both claims hold for the base case $n =1$. Next, if it holds for $n$, then

\begin{align*} \{A_{n+1} = 1 \} &= \{A_n = 1, W_n \leq 0 \} = \{T \geq n, W_n \leq 0 \} = \{T \geq n, S_n \leq 0 \} \\ &= \{T \geq n+1 \}. \end{align*}

Also,

\begin{align*} W_{n+1} = W_n + A_{n+1} X_{n+1} = S_{T\wedge n} + \mathbf{1}_{\{ T \geq n+1 \}} X_{n+1} = S_{T \wedge n+1}, \end{align*}

proving the induction step and hence the claim as desired.

Step 2. By the optional stopping theorem,

$$ \mathbf{E}[W_n] = \mathbf{E}[S_{T \wedge n}] = \mathbf{E}[S_0] = 0. $$

So, if $(W_n)$ has a normal distribution, then we must have $\mathbf{P}(W_n > 0) = \frac{1}{2}$. However,

$$ \mathbf{P}(W_n > 0) = \mathbf{P}(S_{T \wedge n} > 0) = \mathbf{P}(T \leq n) $$

and it is not hard to show that $n \mapsto \mathbf{P}(T \leq n)$ is strictly increasing in $n$ with $\mathbf{P}(T \leq 1) = \frac{1}{2}$. For example, we may argue this by bounding $\mathbf{P}(T = n+1)$ from below by the probability of a specific, tractable scenario in which $T = n+1$ holds:

\begin{align*} \mathbf{P}(T \leq n+1) - \mathbf{P}(T \leq n) &= \mathbf{P}(T = n + 1) \\ &\geq \mathbf{P}(X_1 \in [-1, 0], \ldots, X_n \in [-1, 0], X_{n+1} > n) \\ &> 0 \end{align*}

Consequently, we have $\mathbf{P}(W_n > 0) > \frac{1}{2}$ for all $n \geq 2$ and this proves that $W_n$, $n \geq 2$, cannot have a normal distribution.