$X\sim U(0,1)$, $Y\sim U(0,2)$, how can I find CDF of $T=X+Y$ without knowing the joint PDF of $X$ and $Y$?
Does anyone could help me with this?
$X\sim U(0,1)$, $Y\sim U(0,2)$, how can I find CDF of $T=X+Y$ without knowing the joint PDF of $X$ and $Y$?
Does anyone could help me with this?
QUOTE
$X\sim U(0,1)$, $Y\sim U(0,2)$, how can I find CDF of $T=X+Y$ without knowing the joint PDF of $X$ and $Y$?
END OF QUOTE
If $X\sim U(0,1)$ and $Y=2X$ then $Y\sim U(0,2)$ and $X+Y\sim U(0,3)$, and this has a standard deviation $3$ times as big as that of $X$. But if $X\sim U(0,1)$ and $Y\sim U(0,2)$ and $X$ and $Y$ are independent, then $X+Y$ has a standard deviation only $\sqrt{5}\approx2.236$ times as big as that of $X$ (since $\sqrt{1^2+2^2}=\sqrt 5$). So there is more than one thing that the distribution of $X+Y$ could be, consistenly with the given information about the marginal distributions of $X$ and $Y$.
In other words, you cannot know the distribution of $X+Y$ without further information beyond the two marginal distributions.
There's no real way of tackling this problem without the joint PDF being used indirectly or directly. Luckily, the joint PDF is easy to find because you're given independent random variables; just multiply the two PDFs for $X$ and $Y$ to get the joint PDF.
One you have that, you'll need to do some integration to find the CDF. Be sure to keep track of supports as you do the integration.