Do you know probability ?

Probability Level 4

A point is marked at random on a unit line segment.

Find the expected value of the sum of the squares of the lengths of the two parts.

The answer is 0.667.

1 solution

Brian Charlesworth
Dec 24, 2014

Let the random point on the line segment $[0,1]$ be $x$ . Then the lengths of the two segments are $x$ and $(1 - x)$ . The probable value of the sum of the squares is then

$\displaystyle\int_{0}^{1} (x^{2} + (1 - x)^{2}) dx = \int_{0}^{1} (2x^{2} - 2x + 1) dx = \dfrac{2x^{3}}{3} - x^{2} + x$ ,

which when evaluated from $x = 0$ to $x = 1$ comes out to $\frac{2}{3} = \boxed{0.667}$ to $3$ decimal places.

Nicely done ! :D

Keshav Tiwari - 6 years, 5 months ago

I was scared for a minute that the question had some sort of trick since it was a really simple solution.

Jake Lai - 6 years, 5 months ago

why did you integrate it?

Aryaman Bansal - 6 years, 5 months ago

$x$ is a continuous, random variable with uniform distribution over $[0,1]$ , so integration was the way to go. :)

Brian Charlesworth - 6 years, 5 months ago

so how did we get the most probable answer through it? sorry if I'm asking a basic question but I'm just not able to understand it.

Aryaman Bansal - 6 years, 5 months ago

@Aryaman Bansal – The wording of the question really should be "... find the expected value of the sum of squares ....". The word "probable" is a bit deceiving and not technically correct. The expected value is not necessarily the most probable value but is more of a long-run average, and along with the standard deviation provides a 'profile' of a statistical distribution. This link provides a good introduction to this topic.

Brian Charlesworth - 6 years, 5 months ago

@Brian Charlesworth – In future, if you spot any errors with a problem, you can “report” it by selecting "report problem" in the “dot dot dot” menu in the lower right corner. This will notify the problem creator who can fix the issues.

Calvin Lin Staff - 6 years, 1 month ago

Shouldn't you multiply probability while calculating expected value ?

Abhisek Panigrahi - 3 years, 8 months ago

That is already accounted for in the $\int dx$ , because this is a uniform distribution on length 1.

Can you modify the approach for a line segment of length 10?

Calvin Lin Staff - 3 years, 8 months ago

Is it $\frac{1}{10}\int x^2 + (1 - x)^2 dx$ ?

Abhisek Panigrahi - 3 years, 8 months ago

@Abhisek Panigrahi – You will get the correct value.

For continuous distributions, the expected value is calculated as $E[X] = \int x f(x) \, dx$ . In the case of length 10, the probability density function is $f(x) = \frac{1}{10}$ . So, the "better" way to express it is $E[x^2+(1-x)^2] = \int_1^{10} [x^2 + (1-x)^2] (\frac{ 1}{10}) \, dx$ . ("Better" in the sense of being clear how all the parts fit in, since your first comment indicated that you didn't understand how to "multiply probability" in the continuous case.)

Calvin Lin Staff - 3 years, 8 months ago

@Calvin Lin – Thanks for the explanation. I'm having the doubt of "How the probability density function is $\frac{1}{\text{length of line segment}}$ and not $\frac{1}{\text{no of outcomes} = \infty}$ ?"

Abhisek Panigrahi - 3 years, 8 months ago

@Abhisek Panigrahi – Read up on Continuous Random Variables - Probability Density Function (PDF) .

You are confusing the discrete case with the continuous case (which is a common misconception).

Calvin Lin Staff - 3 years, 8 months ago

Do you know probability ?

The answer is 0.667.

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