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I am stuck on this problem during my review for my stats test.

I know I have to use the convolution formula, and I understand that:

$f_{U_1}(U_1) = 1$ for $0≤U_1≤1$

$f_{U_2}(U_2) = 1$ for $0≤U_2≤1$

but I do not know how to continue on from there. How do I use the convolution formula in this question? Thanks

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    $\begingroup$ what difficulty did you face when you tried the formula? $\endgroup$
    – Siong Thye Goh
    Commented Nov 30, 2018 at 1:45
  • $\begingroup$ I'm not sure how to even set up the equation. I'm guessing I start off with $\int dU_1dU_2$. Is that correct? $\endgroup$
    – peco
    Commented Nov 30, 2018 at 1:58

2 Answers 2

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If you are OK with the convolution formula then it is just a bit of computation. Letting $f_{U_1}(x) = f_{U_2}(x) = 1$ for $x \in [0,1]$ and $0$ elsewhere, and $S = U_1 + U_2$. The convolution of $f_{U_1}$ and $f_{U_2}$ gives us the probability density function of $f_S$, it is $$ f_S(z) = \int_{-\infty}^\infty f_{U_1}(z-x)f_{U_2}(x) \, \mathrm{d}x. $$ Since $f_{U_2}(x) = 1$ on $[0,1]$ we have $$f_S(z) = \int_0^1 f_{U_1}(z-x) \, \mathrm{d}x.$$ We now have to consider cases since $f_{U_1}(z-x) = 1$ only when $ 0\leq z-x \leq 1$, or equivalently when $z-1\leq x\leq z$. Thus, for $0\leq z\leq 1$ we have $$f_S(z) = \int_0^z \, \mathrm{d}x = z.$$ And, for $1\leq z\leq 2$, we have $$f_S(z) = \int_{z-1}^1 \, \mathrm{d}x = 2-z.$$

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  • $\begingroup$ thanks, but I don't understand why we need to consider $1≤z≤2$. I have also seen in the textbook answer that we need to have 2 regions, but is it because we have to consider the region of S which goes from 0 to 2 as $S = U_1 + U_2$? $\endgroup$
    – peco
    Commented Nov 30, 2018 at 2:21
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    $\begingroup$ We need to consider $1\leq z\leq 2$ since it may be the case that $0\leq z-x \leq 1$ and so $f_{U_1}(z-x) = 1.$ As mentioned above, $f_{U_1}(z-x) = 1$ is satisfied whenever $z-1\leq x\leq z$ and note that we are only considering $0\leq x\leq 1$ from the limits of the integral. We need to consider the two regions since both inequalities can be satisfied for each region: for $0\leq z\leq 1$ - the inequalities are satisfied whenever $0\leq x\leq z$; for $1\leq z \leq 2$ - the inequalities are satisfied whenever $z-1\leq x\leq 1$. Hopefully this helps? $\endgroup$
    – Timothy Hedgeworth
    Commented Nov 30, 2018 at 3:05
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Let $s \in [0,2]$

\begin{align} f_S(s) &= \int_{-\infty}^\infty f_{U_1}(t)f_{U_2}(s-t) \, dt \\ &= \int_0^1 1\cdot f_{U_2}(s-t) \, dt \end{align}

Now we just need to integrate over the region when $f_{U_2}$ is positive.

$$0 \le s-t \le 1$$

$$-1 \le t-s \le 0$$

$$s-1 \le t \le s$$

Hence

$$f_S(s) = \int_{\max(0,s-1)}^{\min{(1,s)}} 1 \, dt$$

I will leave the simplication as an exercise.

You might like to consider when $s \ge 1$ and $s<1$.

Remark: In the convoltuion formula, we are integrating only over one variable $t$.

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  • $\begingroup$ sorry for asking so much, but I know min(1,s) means the minimum. But how do you find what it equals to? $\endgroup$
    – peco
    Commented Nov 30, 2018 at 2:16
  • $\begingroup$ If $s< 1$, then we have $\min(1,s)=s$ and if we have $s \ge 1$, then we have $\min(1,s)=1$. $\endgroup$
    – Siong Thye Goh
    Commented Nov 30, 2018 at 2:22

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