STEP I 1997 question 14 solution

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STEP I 1997 question 14 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP 1997 Solutions > STEP I 1997 question 14 solution

f(x) and T(x) are the probability (density) of the maximum height being x, and the cost if the maximum heigh becomes x, respectively. The expected cost, say, C, is then given by

$\displaystyle C = \int_0^\infty f(x)T(x)dx$

$\displaystyle C = \int_0^y e^{-x}ydx + \int_y^\infty e^{-x}(y + r + s(x-y))dx$

$\displaystyle C = \int_0^\infty ye^{-x}dx + \int_y^\infty e^{-x}(r + sx - sy)dx$

$\displaystyle C = \int_0^\infty ye^{-x}dx + \int_y^\infty e^{-x}(r - sy)dx + \int_y^\infty sxe^{-x}dx$

$\displaystyle C = \int_0^\infty ye^{-x}dx + \int_y^\infty e^{-x}(r - sy)dx + \left[-sxe^{-x}\right]_y^\infty + \int_y^\infty se^{-x}dx$

$\displaystyle C = \left[-ye^{-x}\right]_0^\infty + \left[-(r-sy)e^{-x}\right]_y^\infty + \left[-sxe^{-x}\right]_y^\infty + \left[-se^{-x}\right]_y^\infty$

(Note that $\displaystyle \lim_{x\rightarrow\infty} xe^{-x} = 0$ , since $\displaystyle e^x$ grows faster than $\displaystyle x$ )

$\displaystyle C = y + (r-sy)e^{-y} + sye^{-y} + se^{-y}$

$\displaystyle C = y + (r+s)e^{-y}$

Differentiating:

$\displaystyle C' = 1 - (r+s)e^{-y}$

which is strictly increasing. Setting $\displaystyle C' = 0$

$\displaystyle 1 = (r + s)e^{-y}$

$\displaystyle e^y = (r + s)$

we get the minimum at

$\displaystyle \displaytyle y = \ln (r + s)$

as required.

If $\displaystyle (r + s) < 1$ , the minimum will still be at $\displaystyle y = \ln (r + s)$ , only that since $\displaystyle \ln (r + s) < 0$ , the probability that this happens anyway is zero. If y < 0, they won't actually gain anything by lowering the prepared height (it is sensible to assume that they don't get any money back if they prepare for a negative height); the only thing that will happen is that if the actual water height is larger than zero, they will lose out more, because the (X-y) part of the cost is larger. So instead, they should aim at having y = 0. If y becomes larger than that, the derivative of the expected cost,

$\displaystyle C' = 1 - (r + s)e^{-y}$

will always be positive for y>=0. The expected cost is thus increasing for y >= 0, and so the lowest cost they can get is at y = 0. Thus the authorities shouldn't prepare at all when r + s < 1.

Solution by ukgea

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