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$X$ and $Y$ are independent random variables whose marginal PDF's are uniform on $[0,5],$ and $Z=X+Y$. I want to find the Probability Density Function.

To start I assume I need to find the cumulative density function. I tried to follow a similar problem here, but was unable to adapt it to my own problem.

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  • $\begingroup$ The pdf for $Z$ is the convolution of your two pdfs. If you take the convolution of two constants ($=1/5$) you should end up with a triangle-shaped pdf. $\endgroup$
    – GReyes
    Commented Apr 8, 2019 at 0:01
  • $\begingroup$ How do I take a convolution in this context? $\endgroup$
    – Peetrius
    Commented Apr 8, 2019 at 0:02
  • $\begingroup$ What do you know and not know about the definition of convolution. In your case it is very straightforward. $\endgroup$ Commented Apr 8, 2019 at 0:25
  • $\begingroup$ All I know is that I need to take $f_Z(z)=\int_{-\infty}^\infty f_X(x)f_Y(z-x)\,dx.$, but beyond that it is not straightforward for me. $\endgroup$
    – Peetrius
    Commented Apr 8, 2019 at 0:32
  • $\begingroup$ Observe that $f_X=1/5$ in $[0,5]$ and zero elsewhere. Same for $Y$. If you fix some $z$, your integral only extends to those $x$ such that $x\in[0,5]$ and $z-x\in[0,5]$. Look at all possible cases for the intersection. $\endgroup$
    – GReyes
    Commented Apr 8, 2019 at 0:54

1 Answer 1

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If you have not learned about convolution products, just find the CDF first. $$ P(X+Y\leq t)=\int_0^{t-u}\frac{1}{5}(\int_0^{u} \frac{1}{5}du)dx $$ for $0\leq t \leq 5$. It is actually better to draw the square $[0,5]^2$ and evaluate the area geometrically.

The CDF should be $$ \frac{1}{50}t^2, (0\leq t\leq 5)\\ 1-\frac{1}{50}(10-t)^2, (5\leq t\leq 10)\\ 0, (\text{otherwise}) $$

If you differentiate this you get the PDF $$ \frac{1}{25}t, (0\leq t\leq 5)\\ \frac{1}{25}(10-t), (5\leq t\leq 10)\\ 0, (\text{otherwise}) $$ which is a "triangle" if you plot it.

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  • $\begingroup$ Thank you! One question, should there be a minus sign in front of $\frac{1}{25}(10-5)$? $\endgroup$
    – Peetrius
    Commented Apr 8, 2019 at 1:34
  • $\begingroup$ No. PDF cannot be negative. Just be careful with differentiation. $\endgroup$
    – Ma Joad
    Commented Apr 8, 2019 at 1:37
  • $\begingroup$ Fascinating, okay, I will keep that in mind. $\endgroup$
    – Peetrius
    Commented Apr 8, 2019 at 1:40

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