Expectation of a shorter piece

The original question is:

Suppose a 12 inch metal rod inside a factory of some sort is secured at both ends. The rod is then placed under significant pressure until it breaks cleanly. Let XX be the distance from the left end where the break occurs, and the pdf of XX is given as:
fX(x)={(x24)(1−x12),0≤x≤120,otherwisef_X(x)=\left\{\begin{array}{rl}
\left(\dfrac{x}{24}\right)\left(1-\dfrac{x}{12}\right), & 0\le x \le 12 \\
0, & \text{otherwise}
\end{array}\right.
Find the expected length of the shorter segment when the rod breaks.

I found the expected value using:
E[X]=124∫120(x2−x312)dx=6E[X]=\dfrac{1}{24} \int_0^{12}\left({x^2-\dfrac{x^3}{12}}\right)dx=6
But the answer is given as 3.753.75 and I have no idea how they got that. Wouldn’t I expect the rod to break exactly in the middle, and therefore, there wouldn’t be a shorter or longer segment in the first place? Or is this conditional expectation?

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so if the rod breaks into 2 perfectly its 6. what is the probability of it doing this? how many standard deviations off will we be on average?
– shai horowitz
2 days ago

  

 

@shaihorowitz The probability of breaking perfectly is P(X=6)=0P(X=6)=0 since it’s a continuous distribution, but I’m not sure what else you mean. The standard deviation was σX≈2.6834\sigma_X \approx 2.6834.
– nexicon
2 days ago

3

 

Notice that the question looks for E(min{X,12−X})\mathbb{E}(\min\{X,12-X\}) rather than E(X)\mathbb{E}(X).
– Fimpellizieri
2 days ago

  

 

@Fimpellizieri Could you explain how to find the density function for that minimum in this context, please?
– nexicon
2 days ago

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1 Answer
1

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Let Y=min{X,12−X}Y = \min\{X,12-X\}. Observing that 0 \leq X \leq 120 \leq X \leq 12, we have that:

When X\leq 6X\leq 6, Y=XY=X
When X \geq 6X \geq 6, Y=12-XY=12-X

so in other words:

\begin{equation}
Y=\left\{
\begin{array}{lr}
X & ;\,\,X\leq 6\\
12-X & ;\,\,X\geq 6
\end{array}\right.\end{equation}\begin{equation}
Y=\left\{
\begin{array}{lr}
X & ;\,\,X\leq 6\\
12-X & ;\,\,X\geq 6
\end{array}\right.\end{equation}

Hence the integral becomes:

\mathbb{E}(Y)=\int_0^6x\cdot \left(\dfrac{x}{24}\right)\left(1-\dfrac{x}{12}\right)\,dx\,+\,\int_6^{12}(12-x)\cdot \left(\dfrac{x}{24}\right)\left(1-\dfrac{x}{12}\right)\,dx\mathbb{E}(Y)=\int_0^6x\cdot \left(\dfrac{x}{24}\right)\left(1-\dfrac{x}{12}\right)\,dx\,+\,\int_6^{12}(12-x)\cdot \left(\dfrac{x}{24}\right)\left(1-\dfrac{x}{12}\right)\,dx