Find $P(\xi+\eta < t\lambda)$.

Question

Let $\xi$, $\eta$ and $\lambda$ be independent random variables uniformly distributed on the interval $[0,1]$, and let $t$ be a fixed number. Find $P(\xi+\eta < t\lambda)$.

My solution:

1)Let the V random variable $=X+Y.P(X+Y \leq z)= \int_0^z \int_0^1f_x(x)f_y(y-x)dxdy$. Than $P(V<\lambda t)=\int_0^tP(V\leq\lambda_i|\lambda=\lambda_i)f_{\lambda t}(z)dz = P(V\leq\lambda_i)f_\lambda(z)dz =\int_0^t\int_0^z\int_0^1(f_x(x)f_y(y-x)dxdy)f_{\lambda t}(z)dz$

2)$\int_0^z\int_0^1(f_x(x)f_y(y-x)dxdy) = \int_0^z\int_0^z(f_x(x)f_y(y-x)dxdy)$ Because x and y >=0 ,and when x or y >z $P(X+Y\leq z)$=0.

3)We need only positive values f_y(y-x) ,due to symmetry $\int_0^z\int_0^1(f_x(x)f_y(y-x)dxdy) =\frac{\int_0^z\int_0^zdxdy}{2} = \frac{z^2}{2}$

And in the end

4)$\int_0^t\frac{z^2}{2}f_{\lambda t}(z)dz = \frac{t^2}{6}$(It is easy to prove that$f_{\lambda t}(z)=\frac{1}{t} z\in [0, t]$)

I am not sure of the correctness of this decision, or rather in paragraph 3.

I will be glad if you can help me solve this problem.

Answer 1

$X+Y$ follows a triangular distribution with pdf $f_{X+Y}:x\mapsto x 1_{x\in [0,1]} + (2-x)1_{x\in [1,2]}$.

Letting $F$ denote the cdf of $X+Y$, $F$ is the piecewise function $$F:u\mapsto \begin{cases} 0 &\text{if } u\leq 0\\ \frac{u^2}2 &\text{if } u\in [0,1] \\ 1-\frac{(2-u)^2}2 &\text{if } u\in [1,2]\\ 1 &\text{if } u\geq 1. \end{cases}$$

Since $Z$ is independent of $(X,Y)$, $$P(X+Y < tZ) = E[F(tZ)] = \int_0^1F(tz) dz = \frac 1t \int_0^t F(z) dz.$$

The final expression is obtained by substituting the expression of $F$:

$$P(X+Y < tZ) = \begin{cases} 0 &\text{if } t\leq 0 \\ \frac{t^2}6 &\text{if } t\in [0,1] \\ -\frac{t^2}{6}+t+\frac{1}{3 t}-1 &\text{if } t\in [1,2]\\ 1-\frac 1t &\text{if } t\geq 2. \end{cases}$$