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I am trying to solve the following problem

Theo the monkey and Colby the hippo have a 20-ounce cake. Colby cuts it in half and lets Theo pick the larger piece. But Colby is a hippo, so he can’t cut very well. In particular Colby cuts the cake into two pieces, one to his left, and one to his right; the size of the piece to Colby’s right is 10 + U 1 + U 2 , where U 1 and U 2 are independent random variables, each uniformly distributed on the interval [−2,2]. In parts (a) to (c), Theo is a clever monkey and always chooses to take the larger piece of cake. Let T be the size of the piece of cake Theo chooses.

(a) What is the density of T?

The solution is given as below:

(a) The size of the piece to the right has triangular density, supported on [8,12] with a maximum at 10. This is symmetric, so the size of the larger piece is linear on [10,12], decreasing from a maximum at 10 to zero at 12. The equation of the density is 6−t/2 on [10,12] and 0 elsewhere.

Yet this doesn't seem to make sense. If the interval is from $[-2, 2]$ shouldn't the supports be at $[6,14]$? At first, I thought there might have been a mistake with the question and the interval was actually from $[-1,1]$ yet this doesn't work with the answer either, because then the slope of the right side would be $\frac {1}{4}$ and the intercept would be 2.

Is the solution incorrect or am I just missing something?

Community
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Dider
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1 Answers1

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The solution is incorrect as it stands - the two $[-2,2]$ distributions should add to a $[-4,4]$ triangle.

However if $U1$ and $U2$ were uniform on $[-1,1]$, the density equation would be correct, because the two sides of the triangular distribution fold together with Theo taking the bigger piece, so the slope is $-\frac 12,$ not $-\frac 14$. The shortcut is to remember that the area under the density graph must be $1$, so the right triangle sitting on a base of $2$ units will have maximum density value $1$ at measure $10$.

Joffan
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  • Can you elaborate what you mean by "two sides of the triangular distribution fold together "? Do you mean that it's symmetric? – Dider Aug 17 '17 at 22:49
  • I mean that if the cut is at $9.3$, for example, this leads to a $T$ value of $10.7 (= 20-9.3)$, and of course a cut at $10.7$ also leads to the same $T$ value. So the original cut position distribution doubles in weight above ten as it goes to zero below. – Joffan Aug 17 '17 at 23:31