I have looked at many sites with almost enough information for this to make sense... just not quite there yet. So I have this convolution formula (w instead of z and $dx$ instead of $dy$): $$f(w) = \int_{-^\infty}^{\infty} f_x(x)f_y(w-x)dx $$ X and Y are independent, uniform random variables distributed from (0,1) for X and (1,2) for Y
I have a few ideas... $fx(X) = 1_{(0,1)}(x)$ and $fy(Y) = 1_{(1,2)}(y)$ and $f_y(Y) = f_{w-x}(w-x)$
What I am trying to get is the limits. That is, the values $w$ will take on in certain ranges. I am having trouble getting to the pdfs and the appropriate limits. For example: $f_w(w) = what?\;if\; w < somenumber? \;Note: it\;can \;be \lt,\; \gt,\;\le,\;\ge,\;, etc $
And thank you!
*****Edit (twice) So where I am (currently) at is this: $$f(w) = \int_{-^\infty}^{\infty} 1_{(a,b)}(x)f_(x)dx = \int_{0}^{1}f(x)dx $$ $$1_{1,2}(y) = f_{(w-x)}(w-x) = \int_{1}^{2} f_{(w-x)}d(w-x)$$ So, I think I derive the $f(x)dx$, derive $f(y)dy$, (because easier than looking at $(w-x)$), go back and replace the $y=(w-x)$. Then I would have pdfs, I think. Does this appear to be the correct course of action? And thank you again!
**Derived and evaluated:
$$f(x) = \begin{cases} 0, & { x < 0 } \\ 1, & {0 \le x \le 1/2} \\ 0, & {x>1/2} \end{cases}$$
$$f(w-x) = \begin{cases} 0, & {(w-x) < 1}\\ 3/2, & {(w-x) \ge 2}\\ 0, & {(w-x) > 2} \end{cases} $$
So, now I need to work out (from the above) $f_w(w)$.
**Final Answer:
$$f(w) = \begin{cases} 0, & if \;\; {w < 1/2}\\ 1/2, & if\;\; {3/2 \le w \le 2}\\ 3/2, & if\;\; {2 \le w \le 5/2}\\ 2, & if \;\;{w > 5/2} \end{cases} $$
Could someone pretty please check my math?? Thank you! *******@ Joey Doey One more question, from the convolution formula I should be able to get the pdf (also per a comment below) but I'm not sure what to plug in where. I will need to this to find the Expected and Variance of W.
I found this terribly useful and came up with similar answers. Find PDF of $X-Y$
My problem is this part... "You find it from the joint support. $ 0≤z+y≤1,0≤y≤1⟹−1≤z≤1$"
I can't seem to figure out how both of the z+y and y inequalities evaluated to the z inequality. Still working on the pdf of W :( :)
Remember that $X$ is a random variable whereas $x$ is a particular value that it can take. So $f_X$ is a function and $f_X(x)$ a possible value. If $f_X(x)=1_{(0,1)}(x)$, then the following are all equivalent: $$1_{(0,1)}(w-y)=1,$$ $$w-y\in (0,1),$$ $$0<w-y<1.$$ Also note the very important fact that $$\int_{-\infty}^\infty 1_{(a,b)}(x)f(x)\,dx = \int_a^b f(x)\,dx.$$ Hopefully this will get you started.