Recitation: Ambulance Travel Time

What this lecture covers

This short problem finds the distribution of the time it takes an ambulance to travel from its current position to an accident, when both positions are independent and uniformly distributed along a stretch of road. It is a geometric derived-distribution problem: the travel time is a function of two random variables rather than one.

It applies the CDF-then-differentiate method from the derived-distributions lecture to a case involving an absolute difference of two independent uniform random variables, translating the probability into an area calculation.

Key ideas

• Travel time as a function of two variables: with accident location X and ambulance location Y both uniform on [0,l] and independent, travel time is T = |Y - X| / v for a constant speed v.
• Translating the CDF into a region: P(T <= t) becomes the event that Y lies within vt of X, a band-shaped region in the (x,y) square.
• Probability as area under a uniform joint density: since X and Y are independent and uniform, the joint density is a constant, so the probability equals that constant times the area of the band inside the square.
• Area by subtraction: the band's area is found by subtracting the area of two corner triangles (where the band runs off the edge of the square) from the full square.
• Differentiating the CDF piecewise: the resulting CDF is piecewise, and differentiating it in the middle region gives a PDF, while the value is constant (and so has zero density) outside that range.

Walkthrough

Setting up T as a function of X and Y (0:00)

The accident location X and the ambulance's location Y are each uniform on [0,l] and independent. Travel time is T = |Y-X| / v. The event T <= t is rewritten as X - vt <= Y <= X + vt, defining a band-shaped region A in the square [0,l] x [0,l].

Computing the probability as an area (3:12)

Since X and Y are independent and uniform, their joint density is the constant 1/l^2. The probability that T <= t is this constant times the area of region A, which is found by subtracting two triangular corner regions (each with area (1/2)(l - vt)^2) from the full square l^2.

Differentiating to get the PDF (5:16)

Differentiating the resulting CDF with respect to t, for t between 0 and l/v, gives a PDF of 2v/l - 2v^2 t / l^2. Outside that range, the CDF is constant (0 below the range, 1 above it), so the density is 0 there.

Before you watch

• Be comfortable with the CDF-then-differentiate method for finding the distribution of a function of random variables.
• Reviewing how probabilities correspond to areas under a uniform joint density on a square will help with the geometric argument here.

Check your understanding

1. Why is the event T <= t equivalent to Y falling within vt of X?
2. Why does the joint density of X and Y being constant let you replace an integral with an area calculation?
3. Why is the PDF of T equal to 0 for t greater than l/v?