What is the baby-step to derive this?

Question

Could somebody help me with this one quickly? My midterm is coming in 3 hours. I wonder how to go from the first line to the second line? How does (Z of t)^2 suddenly become (Z of t-1) ?

Thank you

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Answer 1

\begin{equation} \begin{aligned} \mathbb{P}(Y_t \leq 0 ) &= \mathbb{P}\left( \frac{1}{\sqrt{2}} (Z_t^2 - 1) \leq 0\right)\\ & = \mathbb{P}\left( \frac{1}{\sqrt{2}} Z_t^2 - \frac{1}{\sqrt{2}} \leq 0\right)\\ & = \mathbb{P}\left( \frac{1}{\sqrt{2}} Z_t^2 \leq \frac{1}{\sqrt{2}}\right)\\ &= \mathbb{P}\left( Z_t^2 \leq 1\right)\\ & = \mathbb{P}\left(Z_{k}^2 \leq 1\right) \end{aligned} \end{equation} where the last equality holds for any $k = 1,2,3,...$. In particular it holds for $k = t-1$. This is due to the i.i.d. assumption on $Z$.

In other words, because $Z_r$ and $Z_s$ are both standard normal random variable (for any values $r$ and $s$) it holds that \begin{equation} \mathbb{P}\left( Z_r^2 \leq 1\right) = \mathbb{P}\left(Z_{s}^2 \leq 1\right). \end{equation}

Answer 2

If the square of number is less than one, the number must lie between $-1$ and $1$, that is, its absolute value is less than one: $z^2<1\iff -1<z<1\iff |z|<1$.