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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 $

$f_Z(x) = (F_X(x - 1)(1/3) + F_X(x - 2)(2/3))'$

$= 1/3*f_X(x-1)+2/3*f_X(x-2)$

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    $\begingroup$ What is the trouble with this? Please show what work and thoughts you have so far, so that people can see where you are having difficulty. $\endgroup$ Apr 6, 2020 at 12:09
  • $\begingroup$ If $Z=X+Y$ then you cannot say $p_Z(z) = p_X(x) + p_Y(y)$ $\endgroup$
    – Henry
    Apr 6, 2020 at 13:00
  • $\begingroup$ @Henry Okay, I tried to attempt the problem a different way $\endgroup$
    – Manny
    Apr 6, 2020 at 15:21

2 Answers 2

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Try to calculate $P(X+Y \le x )$ by conditionning by the value of Y , formally $P(X+Y \le x ) = \sum_{k=0} P(X+k \le x ) P(Y=k)$

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  • $\begingroup$ Hi Kupoc I attempted your suggestion and got to this point not sure how to continue $\endgroup$
    – Manny
    Apr 6, 2020 at 15:21
  • $\begingroup$ Thank you @Kupoc! $\endgroup$
    – Manny
    Apr 6, 2020 at 17:13
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Your edited answer of $$F_Z(x)=P(X + Y \leq x)=F_X(x - 1)(1/3) + F_X(x - 2)(2/3)$$ is correct. But it is a cumulative distribution function, so now take the derivative with respect to $x$ to give the density function $$f_Z(x)=\tfrac13 f_X(x - 1) + \tfrac23 f_X(x - 2)$$

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  • $\begingroup$ Thank you, Henry! $\endgroup$
    – Manny
    Apr 6, 2020 at 17:12

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