# Find the PDF of a Linear Function With Respect to t of Independent Random Processes

The problem goes as follows: "A Stochastic Process $X(t)$ is defined via:

$$X(t,\omega) = A(\omega)t + B(\omega)$$

where $A(\omega) \sim U([-1,1])$ and $B(\omega) \sim U([-1,1])$ are independent random variables.

Determine the first order PDF $f_X(x;t)$ associated with it."

Okay so this is a homework problem and I have been trying to find a similar problem. The difficulty I have faced with $t$ variable introduced in random process. I have found solutions for how to find the PDF via convolution of the sum of two random independent variables (section 4.1 in Intro to Probability by Bertsekas and Tsitsiklis).

One question I have is can I just carry the $t$ variable through while attached to $a$ (where $a$ is a specific member of the set $A$) when finding the joint probability?

Any hints on this would be appreciated.

PS

My attempt to do this (following the book mentioned above):

$$P\left(X(t)<x\mid A=a\right) = P\left(at + B(\omega)<x\mid A=a\right)$$ $$P\left(at + B(\omega)<x\right) = P\left(B(\omega)<x - at\right)$$

then

$$f_{X\mid A}(x\mid at) = f_B(x-at) \Longrightarrow f_{A,X}(at\mid x) = f_A(at)\cdot f_{X\mid A}(x\mid at)$$ $$=f_{A}(at)f_B(x-at)$$

then do the convolution of this.

Honestly this was my first attempt and I am not even sure if I am on the righ track.

Thanks for any help given.

$P\left(X(t)<x\right|A=a) = P\left(at + B(\omega)<x|A=a\right)$

$=P\left(at + B(\omega)<x\right) = P\left(B(\omega)<x - at\right)$

then

$f_{X|A}(x|at) = f_B(x-at)$ => $f_{A,X}(at|x) = f_A(at)*f_{X|A}(x|at)$

$=f_{A}(at)f_B(x-at)$

then do the convolution of this.

Yessish.

\begin{align}f_{X\mid A}(x\mid a;t) ~&=~f_B(x-at) \\ f_{A,X}(a,x;t)~&=~ f_A(a)\,f_B(x-at)\\ f_{X}(x;t)~&=~ \int_\Bbb R f_A(a)\,f_B(x-at)\,\mathrm d a \\ &=~ \tfrac 14\int_\Bbb R \mathbf 1_{a\in[-1;1],x-at\in[-1;1]}\mathrm d a\\ &=~ \tfrac 14\mathbf 1_{x\in[(-t-1);(t+1)]}\int_{\max\{-1,(x-1)/t\}}^{\min\{1,(x+1)/t\}} \mathrm d a &~~~\vdots\end{align}

From here, consider the cases : $[t< -1]$ , $[-1\leqslant t<0]$ , $[0\leqslant t<1]$ , and $[1\leqslant t]$ .

• Okay so I am a bit confused on where to go from here. So the convolution is over the variable $a$, but the effect of $x-at$ on the convolution is confusing to me (since time can vary here). The convolution of two unit step functions is a triangle function (these two are the same size, trapezoid if the widths very), but how does the PDF vary in $x$ and in $t$? I feel like the $t$ must make the triangle function wider or narrower, but I do not know how to show that. I would appreciate a hint if anyone wants to chime in. I do not have a very good intuition for PDFs. – J.Doe Sep 5 '17 at 3:23
• $t$ is a parameter, $x$ is a value; treat them both as constant, but determine the support for the values $x$ give parameter $t$. – Graham Kemp Sep 5 '17 at 3:54