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Suppose $u$ and $v$ are uniform, independent random variables, such that $0 \leq u \leq 2$ and $0 \leq v \leq 3$. Determine the density of $u + v$.

My Approach

Let $x = u + v$. I believe the approach is to break down into three cases, which are

(1) $0 \leq x \leq 2$

(2) $2 \leq x \leq 3$

(3) $3 \leq x \leq 5$

I consider

$$f_U(u) = \left\{\begin{array}{c c}\dfrac{1}{2}, & 0 \leq u \leq 2\\0, & \text{otherwise}\end{array} \right.$$

and

$$f_V(v) = \left\{\begin{array}{c c}\dfrac{1}{3}, & 0 \leq v \leq 3\\0, & \text{otherwise}\end{array} \right.$$

We want to determine $f_{U + V}(x)$. I think that for the first case, my density function is right. I got

$$f_{U + V}(x) = \int_0^2 f_U(u)f_V(x - u)\,du$$

which means that $$f_{U + V}(x) = \int_0^x \dfrac{1}{2}\cdot\dfrac{1}{3}\,du = \dfrac{x}{6}$$

But I'm not sure how to approach the second case. I know that you will need to add the integrand within $0 \leq x \leq 2$ by the integrand within $2 \leq x \leq 3$, but I'm stuck.

However, I'm sure that for the last case, you need to take the difference between 1 and the two density functions at $0 \leq x \leq 2$ and $2 \leq x \leq 3$.

Please offer me suggestions of how to approach the whole problem.

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You have $$\displaystyle f_{U + V}(x) = \int_{\max(0,x-3)}^{\min(2,x)} \dfrac{1}{2}\cdot\dfrac{1}{3}\,du $$ since $U$ cannot be less than $0$ or less than $x-3$, and $U$ cannot be more than $2$ or more than $x$. These become:

  • for $0 \le x \le 2$ you have $\displaystyle f_{U + V}(x) = \int_0^x \dfrac{1}{2}\cdot\dfrac{1}{3}\,du $

  • for $2 \le x \le 3$ you have $\displaystyle f_{U + V}(x) = \int_0^2 \dfrac{1}{2}\cdot\dfrac{1}{3}\,du $

  • for $3 \le x \le 5$ you have $\displaystyle f_{U + V}(x) = \int_{x-3}^2 \dfrac{1}{2}\cdot\dfrac{1}{3}\,du $

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Your formula for $f_{U+V}(x)$ works for all three cases. Just keep going!

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  • $\begingroup$ Good to hear that, but how would one really determine the density function at each interval. This is what I mean $\endgroup$ – NasuSama Nov 14 '13 at 23:59
  • $\begingroup$ For $2 \leq x \leq 3$, if $0 \leq u \leq 2$ then $f_V(x-u) = 1/3$ so the integral is $1/3$. $\endgroup$ – apt1002 Nov 15 '13 at 0:08
  • $\begingroup$ The third case is similar to the first case which you did yourself. $\endgroup$ – apt1002 Nov 15 '13 at 0:09

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