Find one-dimensional distribution function $F(y\mid t)$ of random process $Y(t)$

Question

$ Y(t)=tZ^2;\quad Z\sim U(-2;2); \quad t\ge0. \quad$

I need to

1) find one-dimensional distribution function $F(y|t)$ of random process $Y(t)$.

2) calculate probability that trajectory of the process at moment $t=4$ will intersect interval $[9;25]$.

U is a continuous uniform distribution.

What I've done so far:

1) Probability density function:

$p(y) = \begin{cases} {1 \over 4}, & y\in(-2;2) \\ 0, & y\notin(-2;2) \end{cases} $

Cumulative distribution function:

$F(y) = \begin{cases} 0, & y\le-2 \\ y+2 \over 4, & -2\lt y\lt 2 \\ 1, & y \ge 2 \end{cases} $

$ F(y\mid t)=P(Y(t)\le y)=P(tZ^2 \le y)=P(|Z| \le \sqrt{y \over t})=F_Z( \sqrt{y \over t}) -F_Z( -\sqrt{y \over t})$

$F_Z( \sqrt{y \over t}) = \begin{cases} 0, & \sqrt{y \over t}\le-2 \\ \sqrt{y \over t} +2 \over 4, & -2\lt \sqrt{y \over t}\lt 2 \\ 1, & \sqrt{y \over t} \ge 2 \end{cases} $

$F_Z( \sqrt{y \over t}) = \begin{cases} 0, & \text{no solution} \\ \sqrt{y \over t} +2 \over 4, & 0 \le y \le 4t \\ 1, & y \ge 4t \end{cases} $

$F_Z( -\sqrt{y \over t}) = \begin{cases} 0, & y \ge 4t \\ \sqrt{y \over t} +2 \over 4, & 0\le y\lt 4t \\ 1, & \text{no solution} \end{cases} $

2) $P(4Z^2 \in [9;25])=P(Z^2 \in [{9 \over 4};{25 \over 4}])=P(|Z|\in [{3 \over 2} ; {5 \over 2}])=\cdots$

Can someone explain how to solve the 2nd part? I tried to follow the example I have with two moments and two intervals but I don't seem to understand it either. I can post the example as well if needed. It is solved in two ways.

Edit:

How do I calculate $P(|Z|\in [{3 \over 2} ; {5 \over 2}])$ ?

Answer 1

Compare:

• $Z\sim U(-2;2)$
• How do I calculate $P(|Z|\in [{3 \over 2} ; {5 \over 2}])$ ?

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