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The question:

$A, B\sim \textsf{U}_{[0,1]}$ are the moments two bulbs are ignited, they then are lit for respectively $S,T \sim \textsf{U}_{[0,2]}$. Find the probability two bulbs are lit at the same time.

So far I’ve managed to follow along with the solution: $$P(\text{two bulbs are lit at the same time}) = \underbrace{2}_{\text{symmetry}}P(\{A \leq B \} \cap \{A+S \geq B\} ) $$ which is then turned into the following integral which I do not understand: $$2\int_0^1\int_0^b {2 - (b-a) \over 2} \mathrm{d}b\mathrm{d}a$$ I know that ${2 - (b-a) \over 2}$ is the probability of $S \geq (b-a)$ since $S$’s pdf is $\frac12$, and that for the condition to be fulfilled $S$ needs to be from $(b-a)$ to $2$. what I don’t understand is how is this integral conceived - and how to graph the question since it was my first instinct but I just hadn’t been able to. Can you please fill in a bit of the intermediate steps and show me how to consult a graph in this question please.

kal_elk122
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    I think it should be d$a$d$b$ instead of d$b$d$a$ – Vercingetorix Oct 22 '22 at 18:12
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    Then just consider $\mathbb{R}^2$ with $a$ and $b$ axes (taking the place of $x$ and $y$ respectively). Then the integral limits just represent the area above the $a$ axis and under the line $b = a$ within $0 \leq a \leq 1$. This is just the area (within $0, 1$) where $a \leq b$ and then for each such $(a, b)$ it's weighted by the pdf that $s \geq (b-a)$ as you remarked, and so integrating we get the total probability. – Vercingetorix Oct 22 '22 at 18:15
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    Do you now that the pdf of the sum of two uniform RV (either $A+S$ or $B+T$) is a function with trapezoidal shape ? – Jean Marie Oct 22 '22 at 18:28
  • no - i hadn’t - so basically based on my knowledge that wasn’t the way to go I see – kal_elk122 Oct 22 '22 at 18:29

2 Answers2

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\begin{align} &P(\{A \le B\} \cap \{A + S \ge B \}) \\ &= \int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty \mathbf{1}\{a \le b\} \mathbf{1}\{s \ge b - a\} f_S(s) f_B(b) f_A(a) \, ds \, da \, db \\ &= \frac{1}{2} \int_0^1 \int_0^1 \int_0^2 \mathbf{1}\{a \le b\} \mathbf{1}\{s \ge b - a\} \, ds \, da \, db \\ &= \frac{1}{2} \int_0^1 \int_0^1 \mathbf{1}\{a \le b\} \left(\int_0^2 \mathbf{1}\{s \ge b - a\} \, ds\right) \, da \, db \\ &=\frac{1}{2} \int_0^1 \int_0^1 \mathbf{1}\{a \le b\} \left(\int_{b-a}^2 \, ds\right) \, da \, db \\ &= \frac{1}{2} \int_0^1 \int_0^1 \mathbf{1}\{a \le b\} (2 - (b-a)) \, da \, db \\ &= \frac{1}{2} \int_0^1 \int_0^b (2 - (b-a)) \, da \, db. \end{align}

angryavian
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I would like to show a graphical proof.

The sum $E$ (initial of Endpoint) of 2 uniform random variables $A \sim U_{[0,1]}$ and $S \sim U_{[0,2]}$ is known to be this trapezoidal-shape function:

$$g(x)=\begin{cases} \ \ \ \frac12x&(0\le x\le 1)\\ \ \ \ \frac12&(1\le x\le 2)\\ -\frac12x+\frac32&(2\le x\le 3) \end{cases}\tag{1}$$

This is for interval $[A,\underbrace{A+S}_E]$. Same distribution for interval $[B,\underbrace{B+T}_F]$. Let us consider the complementary event : "the bulbs are not lit at the same time". It is clear, by symmetry, that the probability of this event is twice the probability of having $E<B$.

By (assumed) independence, the joint pdf if the product of the individual pdfs:

$$h(x,y):=f(x)g(y)\tag{2}$$

(where $f$ is the pdf of $U_{[0,1]}$ and $g$ defined by (1)).

$P(E<A)$ is equal to the volume $\frac{1}{12}$ under the surface with equation (2) and above the triangular area $PQR$ (for which $y<x$) shown in the figure below (i.e., the volume of a pyramid with basis area $\frac12$ and height $\frac12$). Doubling this volume gives the final result for the complementary event: $\frac{1}{6}$.

enter image description here

Jean Marie
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