PDF and CDF of Sum of 2 dice 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

How many ways are there to get a sum of $2$? We must get $1$ and then $1$ again with probability $\frac{1}{36}$.
How many ways are there to get a sum of $3$? We must either get a $1$ and then $2$ or vice versa with probability $\frac{2}{36}$. 
How many ways are there to get a sum of $4$? We must either get a $1$ and then $3$, a $3$ and then $1$, or a $2$ and a $2$, giving a probability of $\frac{3}{36}$
Carrying on in this manner, we find that the probabilities appear to follow triangular shape. As commented, letting $X$ be the sum of the the two die, $X$ is a discrete random variable so we need a pmf.
$$ p_{X}(x)=  
\begin{cases} 
\frac{1}{36} & x = 12 \\
\frac{2}{36} & x=11 \\
\frac{3}{36} & x=10 \\
\frac{4}{36} & x=9 \\
\frac{5}{36} & x=8 \\
\frac{6}{36} & x=7 \\
\frac{5}{36} & x=6 \\
\frac{4}{36} & x=5 \\
\frac{3}{36} & x=4 \\
\frac{2}{36} & x=3 \\
\frac{1}{36} & x=2 \\
0 & \text{otherwise} \\
\end{cases} $$
If you wish to construct a probability density function, despite it not making much sense in this situation, you could so something like
$$ f_{X}(x)=  
\begin{cases} 
\frac{1}{36} & x \geq 12 \\
\frac{2}{36} & 11 \leq x \lt 12 \\
\frac{3}{36} & 10 \leq x \lt 11 \\
\frac{4}{36} & 9 \leq x \lt 10 \\
\frac{5}{36} & 8 \leq x \lt 9 \\
\frac{6}{36} & 7 \leq x \lt 8 \\
\frac{5}{36} & 6 \leq x \lt 7 \\
\frac{4}{36} & 5 \leq x \lt 6 \\
\frac{3}{36} & 4 \leq x \lt 5 \\
\frac{2}{36} & 3 \leq x \lt 4 \\
\frac{1}{36} & 2 \leq x \lt 3 \\
0 & x \lt 2 \\
\end{cases} $$
And the cumulative density function would be something like
$$ F_{X}(x)=  
\begin{cases} 
1 & x \geq 12 \\
\frac{35}{36} & 11 \leq x \lt 12 \\
\frac{33}{36} & 10 \leq x \lt 11 \\
\frac{30}{36} & 9 \leq x \lt 10 \\
\frac{26}{36} & 8 \leq x \lt 9 \\
\frac{21}{36} & 7 \leq x \lt 8 \\
\frac{15}{36} & 6 \leq x \lt 7 \\
\frac{10}{36} & 5 \leq x \lt 6 \\
\frac{6}{36} & 4 \leq x \lt 5 \\
\frac{3}{36} & 3 \leq x \lt 4 \\
\frac{1}{36} & 2 \leq x \lt 3 \\
0 & x \lt 2 \\
\end{cases} $$
