High School Continuous uniform distribution - expected values

[thebosonbreaker]

TL;DR

If the area of a square is uniformly distributed over an interval, why is the expected value of the length different to the square root of the expected value of the square's area?

Hello,
I am currently stumped over a question that has to do with the continuous uniform distribution. The question was taken from a stats exam, and while I understand the solution given in the mark scheme, I don't understand why my way of thinking doesn't work.

The problem is:
The sides of a square are of length L cm and its area is A cm^2. Given that A is uniformly distributed on the interval [15, 20], find E(L).

The mark scheme solution uses integration [ by writing L = A^(1/2) ] so that E(L) = ∫(a^(1/2)) x (1/5)da between 15 and 20. I appreciate that this is making use of the fact that E[g(x)] = ∫g(x)f(x) dx for a CRV.

On the other hand, why can't we simply find E(A) and take its square root? If we expect the area to be E(A), is the length when it has this area not equal to the expected length?

If somebody could clarify this I would be very grateful for your help.

PS it just so happens in this case that both methods give you the same answer, but I know that my way is wrong and the mark scheme won't allow it.

 

[PeroK]

Summary: If the area of a square is uniformly distributed over an interval, why is the expected value of the length different to the square root of the expected value of the square's area?

Hello,
I am currently stumped over a question that has to do with the continuous uniform distribution. The question was taken from a stats exam, and while I understand the solution given in the mark scheme, I don't understand why my way of thinking doesn't work.

The problem is:
The sides of a square are of length L cm and its area is A cm^2. Given that A is uniformly distributed on the interval [15, 20], find E(L).

The mark scheme solution uses integration [ by writing L = A^(1/2) ] so that E(L) = ∫(a^(1/2)) x (1/5)da between 15 and 20. I appreciate that this is making use of the fact that E[g(x)] = ∫g(x)f(x) dx for a CRV.

On the other hand, why can't we simply find E(A) and take its square root? If we expect the area to be E(A), is the length when it has this area not equal to the expected length?

If somebody could clarify this I would be very grateful for your help.

PS it just so happens in this case that both methods give you the same answer, but I know that my way is wrong and the mark scheme won't allow it.

You could look at it the other way round. Suppose the length was uniformly distributed, would that give a uniformly distributed area?

To simplify things you could further consider a simple discrete case.

If the length is $1, 2$ or $3$ with equal probability, then the expected length is $2$.

But, the area is $1, 4$ or $9$ with equal probability, giving an expected area of $14/3$.

 

[PeroK]

PS it just so happens in this case that both methods give you the same answer, but I know that my way is wrong and the mark scheme won't allow it.

PS You may approximately get the same answer in this case , but it shouldn't be the same. From that point of view it's a poor question, as the two answers are nearly the same. This is because it's a relatively small interval relative to the starting point.

 

[Stephen Tashi]

On the other hand, why can't we simply find E(A) and take its square root? If we expect the area to be E(A), is the length when it has this area not equal to the expected length?

It may help your intuition to state it as a generality.

If $X$ is a random variable with expected value $\mu_X$ and $Y = f(X)$ is a function of $X$ then your intuition is that the expected value of $Y$ should be $f(\mu_X)$.

This works when $f$ is a linear function of $X$. On the the other hand consider the example of income tax where there are various "brackets" instead of a flat rate. Is the average tax paid by a citizen equal to the tax paid by a citizen who has the average income?