Working on a problem on probability and I just cannot get myself to make any progress.
Supposed that the 2-dimensional random vector $X=(X_1, X_2)$ has joint p.d.f $f(x_1,x_2)$, i.e. $f$ is a non-negative Borel function s.t. $\int fdy=1$, and for any $x \in \mathbb R^2$, $$P(X\leq x)=\int 1_{(-\infty ,x]}(y)f(y)dy$$ where $1_{(-\infty ,x]}(y)$ is the indicator function. Here $dy$ denotes the integration w.r.t $m_2$, the 2-dimensional Lebesgue measure.
Suppose $X$ is uniformly distributed on $[0,2]^2$, $$f(x)=\frac{1}{4} 1_{[0,2]}(x_1)1_{[0,2]}(x_2)$$ Find the distribution function and probability density function of $Y=X_1+X_2$.
For this question, if I know the distribution function of $X=(X_1,X_2)$, how can I split up the function coordinate wise? Ultimately, are the d.f. and p.d.f. of $Y=X_1+X_2$ simply the sum of the distribution functions and the density functions?
I've been stuck on these problems for a very long time, and I cannot make any progress. Some help would be greatly appreciated.
Best Answer
You are given "$X$ is uniformly distributed on $[0,2]^2$", so joint PDF is $$f_{X_1,X_2}(x_1,x_2)=\frac{1}{4}$$ inside $[0,2]^2$, zero outside. Now, $$f_{X_1}(x_1)=\int f_{X_1,X_2}(x_1,x_2) dx_2=\int_0^2 \frac{1}{4} dx_2=\frac{1}{2}$$ inside $[0,2]$, zero outside. The same is true for $f_{X_2}(x_2)$.
Now we see that $$f_{X_1,X_2}(x_1,x_2)=f_{X_1}(x_1)f_{X_2}(x_2)$$ so $X_1$ and $X_2$ are independent, and the PDF of $Y=X_1+X_2$ is convolution of $f_{X_1}(x_1)$ and $f_{X_2}(x_2)$.