Finding the Expected Value of Stick Breakage

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To find the expected value of the lengths of two pieces from a stick broken at a random point, define a random variable X representing the break point from one end of the 15-inch stick. The lengths of the two pieces can be expressed as L = X and R = 15 - X. The expected values E[L] and E[R] can then be calculated using the uniform distribution of X. The solution involves determining E[X] and applying it to find the expected lengths of both pieces. This approach leads to a clear understanding of the expected values in this stick breakage scenario.
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Homework Statement



Suppose that a point is chosen at random on a stick that has length 15 inches, and that the stick is broken into two pieces at that point. Find the expected value of the lengths of the two pieces.


Homework Equations



E(x)=\sumf(x)xdx from -infinity to +infinity (continuous case)
E(x)=\sumf(x)x for all x (discrete case)

The Attempt at a Solution



X1+X2=15

PLEASE HELP!
 
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any ideas?

try starting from a single uniform random variable X, with uniform distribution, representing break distance from one end...
 
Last edited:
Start by defining a random variable X which represents the point where the stick is broken, measured from the left edge. What is E[X]?

Now define two random variables L and R, where L = length of the left piece, and R = length of the right piece.

Express L and R as functions of X. Then use that result to express E[L] and E[R] as functions of E[X].
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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