What is the probability that $X$ and $Y$ are within 0.1 of each other given a uniformly distributed joint pdf.

Question

Two construction companies make bids of $X$ and $Y$ (in \$$100,000$'s) on a remodeling project. The joint pdf of $X$ and $Y$ is uniform on the space $2<x<2.5,2<y<2.3$. If $X$ and $Y$ are within 0.1 of each other, the companies will be asked to rebid; otherwise, the low bidder will be awarded the contract. What is the probability that they will be asked to rebid?

The uniform pdfs for each variable are $f_x(x) = 2$ and $f_y(y) = \frac{10}{3}$. I assume these variables are independent and thus $f(x,y)=20/3$. We are interested in when the two variables are within 0.1 of each other so $X\le 2.4$. Also we need $|X-Y| \leq 0.1$ We can write this as:

$$A = P(X \le 2.4)$$ $$P(|X-Y| \le 0.1) \mid A)$$

I am not sure how to proceed.

Answer 1

Outline: Draw the rectangle $R$ on which the joint density "lives." Draw the lines $y=x+0.1$ and $y=x-0.1$. We want the probability that $(X,Y)$ lands in the part of $R$ that lies between these two lines. Call this region $S$. The required probability is the area of $S$ divided by the area of $R$.

Answer 2

Although the best way to solve the exercise is to draw a shape - not of the function $f(x,y)$ but of the area $|X-Y|\le 0.1$ - one equivalent way to proceed is the following: $$P(|X-Y|\le 0.1)=P(-0.1\le X-Y\le 0.1)=P(Y-0.1\le X\le Y+0.1)$$ but also $$2\le X\le 2.5$$ and therefore $$\max\{2, Y-0.1\} \le X \le \min \{2.5, Y+1\} \iff \max\{2, Y-0.1\}\le X \le Y+0.1$$ Obviously $$\max\{2, Y-0.1\}=\begin{cases}2, & Y\le 2.1\\ Y-0.1, & Y>2.1\end{cases}$$ Thus the integration limits are $$\int_{2}^{2.1}\int_{2}^{y+0.1}f(x,y)\,dx\,dy+\int_{2.1}^{2.3}\int_{y-0.1}^{y+0.1}f(x,y)\,dx\,dy$$

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Your thought of taking conditional probability is not correct. You do not need to stipulate that $X<2.4$. If you take the integration limits correctly this will be automatically resolved. So, do not take conditional probability.