Median of a symmetric distribution

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iocal

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Median of a distribution(Not necessarily symmetric)

Hi all, new day new question.

Consider the following exercise. Let X be a random variable of the continuous type that has pdf f(x). If m is the unique median of the distribution of X and be is a real constant, show that:

E(|X-b|) = E(|X-m|) + \(\displaystyle 2\int_m^b (b-x) f(x) dx\)

This what I have done so far and that is not much:

\(\displaystyle \displaystyle \mathrm E(|X-b|)= \int_{-\infty}^0 (b-X)\ f(x)\ dx +\int_0^{\infty} (X-b) f(x) dx\)

\(\displaystyle \displaystyle = 2\int_0^{\infty} (X-b)\ f(x)\ dx\)

\(\displaystyle \displaystyle = 2\int_{-\infty}^0 (b-X)\ f(x)\ dx\)

I know that the median is defined as \(\displaystyle \displaystyle \int_{-\infty}^m f(x)\ dx = \int_m^{\infty} f(x)\ dx = 1/2\) but other than that I find myself at a loss.
I don't need the solution, I merely need a starting point.
Thank you.
 
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You title this "symmetric distribution" but don't specify that the distribution is symmetric in the body. Is it?

Also you appear to be assuming that the distribution is symmetric about x= 0. Is that given? If the distribution is symmetric, then "mean" and "median" are the same and the distribution is symmetric about the median.
 
HallsofIvy said:
You title this "symmetric distribution" but don't specify that the distribution is symmetric in the body. Is it?

Also you appear to be assuming that the distribution is symmetric about x= 0. Is that given? If the distribution is symmetric, then "mean" and "median" are the same and the distribution is symmetric about the median.
Click to expand...

You are right, I corrected it. The exercise does not specify anything else. I merely split the integral into two parts then, you can do that since this is an absolute value problem, right?
 
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iocal said:
Hi all, new day new question.

Consider the following exercise. Let X be a random variable of the continuous type that has pdf f(x). If m is the unique median of the distribution of X and be is a real constant, show that:

E(|X-b|) = E(|X-m|) + \(\displaystyle 2\int_m^b (b-x) f(x) dx\)

This what I have done so far and that is not much:

\(\displaystyle \displaystyle \mathrm E(|X-b|)= \int_{-\infty}^0 (b-X)\ f(x)\ dx +\int_0^{\infty} (X-b) f(x) dx\)
[omitted next two lines - not correct.]

I know that the median is defined as \(\displaystyle \displaystyle \int_{-\infty}^m f(x)\ dx = \int_m^{\infty} f(x)\ dx = 1/2\) but other than that I find myself at a loss.
I don't need the solution, I merely need a starting point.
Thank you.
Click to expand...
You must split the integral where the absolute value is 0, namely at x=b. Also, what is the difference between X and x? At least in the integrands, the variable has to be x.

\(\displaystyle \displaystyle \mathrm E(|X-b|)= \int_{-\infty}^b (b-x)\ f(x)\ dx +\int_b^{\infty} (x-b) f(x) dx\)

You will have to treat separately the cases b>m and b<m, because one or the other of your two integrals has to be further split at m. If b<m, then

\(\displaystyle \displaystyle \mathrm E(|X-b|)= \int_{-\infty}^b (b-x)\ f(x)\ dx +\int_b^m (x-b) f(x) dx+\int_m^{\infty} (x-b) f(x) dx\)

I don't know if it is better to work with E[|X-b|] or E[|X-m|]. Probably need both eventually. Perhaps both integrals shwould be split at m, regardless of b < or > m:

\(\displaystyle \displaystyle \mathrm E(|X-b|)= \int_{-\infty}^m (b-x)\ f(x)\ dx +\int_m^b (b-x)\ f(x)\ dx +\int_b^m (x-b) f(x) dx+\int_m^{\infty} (x-b) f(x) dx\)
 
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DrPhil said:
You must split the integral where the absolute value is 0, namely at x=b. Also, what is the difference between X and x? At least in the integrands, the variable has to be x.

\(\displaystyle \displaystyle \mathrm E(|X-b|)= \int_{-\infty}^b (b-x)\ f(x)\ dx +\int_b^{\infty} (x-b) f(x) dx\)

You will have to treat separately the cases b>m and b<m, because one or the other of your two integrals has to be further split at m. If b<m, then

\(\displaystyle \displaystyle \mathrm E(|X-b|)= \int_{-\infty}^b (b-x)\ f(x)\ dx +\int_b^m (x-b) f(x) dx+\int_m^{\infty} (x-b) f(x) dx\)

I don't know if it is better to work with E[|X-b|] or E[|X-m|]. Probably need both eventually. Perhaps both integrals shwould be split at m, regardless of b < or > m:

\(\displaystyle \displaystyle \mathrm E(|X-b|)= \int_-{\infty}^m (b-x)\ f(x)\ dx +\int_m^b (b-x)\ f(x)\ dx +\int_b^m (x-b) f(x) dx+\int_m^{\infty} (x-b) f(x) dx\)
Click to expand...


Okay, thanks.
We have \(\displaystyle \displaystyle \mathrm E(|X-m|)=m \int_{-\infty}^m f(x)\ dx -\int_{-\infty}^m x\ f(x)\ dx +\int_m^{\infty} x\ f(x)\ dx -m\int_m^{\infty} f(x)\ dx\)

\(\displaystyle \displaystyle = -\int_{-\infty}^m x\ f(x)\ dx + \int_m^{\infty} x\ f(x) \ dx = 2\int_m^{\infty} x\ f(x)\ dx\ because \int_{-\infty}^m f(x)\ dx = \int_m^{\infty} f(x)\ dx = 1/2\)


Then simplifying the expression for E(|X-b|) similarly we arrive at the equality. Did I get it right?
 
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iocal said:
Okay, thanks.
We have \(\displaystyle \displaystyle \mathrm E(|X-m|)=m \int_{-\infty}^m f(x)\ dx -\int_{-\infty}^m x\ f(x)\ dx +\int_m^{\infty} x\ f(x)\ dx -m\int_m^{\infty} f(x)\ dx\)

\(\displaystyle \displaystyle = -\int_{-\infty}^m x\ f(x)\ dx + \int_m^{\infty} x\ f(x) \ dx = 2\int_m^{\infty} x\ f(x)\ dx\ \) because \(\displaystyle \int_{-\infty}^m f(x)\ dx = \int_m^{\infty} f(x)\ dx = 1/2\)


Then simplifying the expression for E(|X-b|) similarly we arrive at the equality. Did I get it right?
Click to expand...
One more relevant step - you have to recognize (or prove) that

\(\displaystyle \displaystyle \int_b^\infty dx - \int_m^\infty dx = \int_b^m dx\)
 
DrPhil said:
One more relevant step - you have to recognize (or prove) that

\(\displaystyle \displaystyle \int_b^\infty dx - \int_m^\infty dx = \int_b^m dx\)
Click to expand...

Thank you, I believe you can do that with the CDF.

\(\displaystyle \displaystyle \int_b^{\infty} dx -\int_m^\infty dx = 1-F(b)-(1-F(m)= F(m)-F(b) = \int_b^m dx\)

Again, thanks a lot man.
 
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