Please provide all work so I understand. A motorist has a breakdown at a random

Question

Please provide all work so I understand.

A motorist has a breakdown at a random point on the CT Turnpike, checks GPS, and walks to the closer end of the Turnpike. Of interest is the expected walking distance. Re-formulation for analysis (assuming the Turnpike is of length L): suppose that X is uniformly distributed on [0, L] What is E [min(X, 1 - X)] ? notation: for any number x, min(x, 1 - x) signifies the smaller of the two values x or 1 - x (and 7) if X = 1/2). To get a feeling for the function min(a:, 1 - x), graph it on the interval [0, 1] and find (different) formulas for it on [0, 1/2] and [1/2, 1].

Explanation / Answer

Solution

Back-up Theory

If a continuous random variable, X, is uniformly distributed over the interval (a, b), then the pdf (probability density function) of X is given by

f(x) = 1/(b – a) …………….…………………………………………………………(1)

CDF (cumulative distribution function) = P(X t) = (t – a)/(b – a)………………….(2)

If a continuous random variable, X, has pdf (probability density function) f(x), for a x b, and 0 otherwise, then, Mean (average) of X = E(X) = integral from ‘a’ to ‘b’ of {x.f(x)} ……………...…. (3)

Min{x, 1 – x} over a range (0. L) = x, for 0 x L/2

= 1- x, for L/2 x L……………………………(4)

Now, to work out the solution,

Let X = walking distance. Then, X ~ Uniform (0, L) [given] and hence [vide (1) under Back-up Theory],

f(x) = 1/L………………………………………………………….(5)

Let for convenience in explaining and presentation, y = [min(x, 1 – x)]. Then,

[vide (4) under Back-up Theory], y = x, for 0 x L/2

= 1- x, for L/2 x L

Now, [vide (3) under Back-up Theory],

E[min(x, 1 – x)] = E(Y)

= integral from 0 to L of {y.p(y)}

Since f(y) takes two different values, the above integral must be expressed as a sum of two integrals as follows:

integral from 0 to L of {y.p(y)}

= integral from 0 to L/2 of {y.p(y)} + integral from L/2 to L of {y.p(y)}

= integral from 0 to L/2 of {x.(1/L)} + integral from L/2 to L of {(1 - x).(1/L)} [vide (5) above]

= integral from 0 to L/2 of {x/L)} + integral from L/2 to L of {(1 - x)/L}

= (1/L)[(x2/2)(0, L/2) + {x - (x2/2)}(L/2, L)}] = (1/L){(L/2) - (L2/8)}

= {(½) - (L/8)} ANSWER

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