i'm trying to obtain the PDF and CDF of the sum of 2 dice toss.
There are tons of elementary exercise where is asked to find the exact probability,
but what about the PDF and CDF?
i thought this is a convolution of discrete uniform PDF. but i don't know where to start to find it.
it's like $P(X_1+X_2<y)=$ but from here i don't know how to continue… i can't substitute $1/n=X_1$ and $1/n=X_1$
this would make nosense.
Thank You
Best Answer
Clearly the "exact probability of the sum" is the PDF.
$\mathsf P(X_1{+}X_2{=}y)=\dfrac 1{36}\times\begin{cases}(y-1) &:& y\in\{2,3,4,5,6,7\}\\[1ex] (13-y) &:& y\in\{8,9,10,11,12\}\\0 &:& \textsf{else}\end{cases}$
And the accumulation of this is the CDF.
$\mathsf P(X_1{+}X_2{\leq}y)=\dfrac 1{72}\times\begin{cases}0&:& y\leq 1\\[1ex]\lfloor y\rfloor(\lfloor y\rfloor-1) &:& \lfloor y\rfloor\in\{2,3,4,5,6,7\}\\[1ex] 25\lfloor y\rfloor-\lfloor y\rfloor^2-84 &:& \lfloor y\rfloor\in\{8,9,10,11\}\\[1ex] 72 &:& y\geq 12 \end{cases}$