Let X and Y be independent random variables. The random variable X has probability density function p(x) and Y is a discrete random variable having just two values: 1 with probability 1/3 and 2 with probability 2/3.
Since Z is the sum of two random variables it is known that the sum of two probability distributions is another probability distribution. To calculate:
$P(X + Y \leq x)$
$= \sum_{k=1}^2P(X + k \leq x)P(Y = k) = P(X + 1 \leq x)P(Y = 1) + P(X + 2 \leq x)P(Y = 2)$
$=P(X \leq x – 1)(1/3) +P(X \leq x – 2)(2/3)$
$=F_X(x – 1)(1/3) + F_X(x – 2)(2/3)$
Now, to find the pdf of one must take the derivative of the CDF with respect to x $\implies $
$f_Z(x) = (F_X(x – 1)(1/3) + F_X(x – 2)(2/3))'$
$= 1/3*f_X(x-1)+2/3*f_X(x-2)$
Best Answer
Try to calculate $P(X+Y \le x )$ by conditionning by the value of Y , formally $P(X+Y \le x ) = \sum_{k=0} P(X+k \le x ) P(Y=k)$