Cutting a 1 metre stick, what is the expected length?

Question

I am unable to understand 2 things in this problem:

1. Why is $f_X (x)$ = 1, should it not be 1/x because every point has the same chance of being cut. Does that mean every point has P=1 of being cut? However, the cdf on [0,1] seems to be 1, which is fine. edit: Nvm I figured it out, 1/x where x=1m is 1...

2. How come the expected value is not 0.5m ? If every point has the same probability of being cut, then it is reasonable to choose the mean, which is 0.5m. However, the answer is 0.75. I am not sure if that is because the question asks for "the length of the longer piece".

Answer 1

1) The density probability function being a constant is the telltale sign that the distribution is uniform over whatever interval is considered. Being constant means that each point has the same probability (density) to be chosen, which is intended. The value of the constant is then determined to make sure (as you wrote) that the intergral overt the whole considered interval is $1$.

2) The expected value for say the first piece is actually $0.5m$. The key is (as you write) that it asks for the longer piece. The longer piece must be at least $0.5m$ long, because if it was shorter, than the other piece would be shorter still, and together they can't be $1m$ long in their sum.

Answer 2

The longer piece has a length uniform in $[0.5,1]$ (and the shorter uniform in $[0,0.5]$).

The expectation of the longer length is the middle of the range,

$$\frac{0.5+1}{2}.$$