Symmetric density

Hi all,

It seems there is something I am missing here and I would appreciate some input. The exercise is as follows:

Let a RV X of the continuous type have pdf f(x) whose graph is symmetric with respect to x=c. If the mean value of X exists, show that E(X)=c

Hint: Show that E(X-c) equals zero by writing E(X-c) as the sum of two integrals: one from minus infinity to c and the other from c to infinity. In the first let y=c-x and in the second z=x-c
Finally use the symmetry condition f(c-y) = f(c+y) in the first

Ok so after having made the appropriate substitutions we have:

∫yf(c+y)dy + ∫zf(z+c)dz

the first integral goes from -infinity to c and the second from c to infinity. I am sorry for that, you also have to tell me how to type math equations, it is very annoying to say the least.

We have to prove that this equation goes to zero but how do we do that? This is very easy conceptually, of course a pdf that is symmetric to c has expected value of c!
But I am certain there is something I am missing here. Could you please point me to that direction?

iocal said:

∫yf(c+y)dy + ∫zf(z+c)dz

the first integral goes from -infinity to c and the second from c to infinity. I am sorry for that, you also have to tell me how to type math equations, it is very annoying to say the least.

Click to expand...

Demonstration of LaTeX. Right-click on the following equation, then select Show Math As > TeX Commands

\(\displaystyle \displaystyle \int_{- \infty}^c y\ f(c+y)\ dy + \int_c^{\infty}z\ f(z+c)\ dz \)

Cut&Paste the resulting text into a text file (e.g., Notepad) so you will have an example to look at. Enclosing this text in a block followed by [/tex] and preceded by \(\displaystyle will generate the equation. I learned by looking at other peoples equations this way, and from this web site:

www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols\)

DrPhil said:

Demonstration of LaTeX. Right-click on the following equation, then select Show Math As > TeX Commands

\(\displaystyle \displaystyle \int_{- \infty}^c y\ f(c+y)\ dy + \int_c^{\infty}z\ f(z+c)\ dz \)

Cut&Paste the resulting text into a text file (e.g., Notepad) so you will have an example to look at. Enclosing this text in a block followed by [/tex] and preceded by \(\displaystyle will generate the equation. I learned by looking at other peoples equations this way, and from this web site:

www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols\)

Click to expand...

\(\displaystyle


Thank you, and please feel free to look at the exercise too!\)

iocal said:

Hi all,

It seems there is something I am missing here and I would appreciate some input. The exercise is as follows:

Let a RV X of the continuous type have pdf f(x) whose graph is symmetric with respect to x=c. If the mean value of X exists, show that E(X)=c

Hint: Show that E(X-c) equals zero by writing E(X-c) as the sum of two integrals: one from minus infinity to c and the other from c to infinity. In the first let y=c-x and in the second z=x-c
Finally use the symmetry condition f(c-y) = f(c+y) in the first

Ok so after having made the appropriate substitutions we have:

\(\displaystyle \displaystyle \int_{- \infty}^c y\ f(c+y)\ dy + \int_c^{\infty}z\ f(z+c)\ dz\)

We have to prove that this equation goes to zero but how do we do that? This is very easy conceptually, of course a pdf that is symmetric to c has expected value of c!
But I am certain there is something I am missing here. Could you please point me to that direction?

Click to expand...

I need to back up to be sure I understand the substitutions. I hope you remembered that a substitution also changes the limits of a definite integral.

\(\displaystyle \displaystyle \mathrm E[x-c] = \int_{- \infty}^c (x-c)\ f(x-c)\ dx + \int_c^{\infty}(x-c)\ f(x-c)\ dx\)

.............\(\displaystyle \displaystyle = \int_{- \infty}^0 (-y)\ f(c-y)\ (-dy) + \int_0^{\infty}z\ f(z+c)\ dz\)

.............\(\displaystyle \displaystyle = -\int_0^{\infty} y\ f(c+y)\ dy + \int_0^{\infty}z\ f(z+c)\ dz\)


Hmmm. Maybe you didn't substitute the limits...

DrPhil said:

I need to back up to be sure I understand the substitutions. I hope you remembered that a substitution also changes the limits of a definite integral.

\(\displaystyle \displaystyle \mathrm E[x-c] = \int_{- \infty}^c (x-c)\ f(x-c)\ dx + \int_c^{\infty}(x-c)\ f(x-c)\ dx\)

.............\(\displaystyle \displaystyle = \int_{- \infty}^0 (-y)\ f(c-y)\ (-dy) + \int_0^{\infty}z\ f(z+c)\ dz\)

.............\(\displaystyle \displaystyle = -\int_0^{\infty} y\ f(c+y)\ dy + \int_0^{\infty}z\ f(z+c)\ dz\)


Hmmm. Maybe you didn't substitute the limits...

Click to expand...

Oh **** it. Apologies for my french and for like the hundrenth time, thank you.

And well then the final equation \(\displaystyle \displaystyle = -\int_0^{\infty} y\ f(c+y)\ dy + \int_0^{\infty}z\ f(z+c)\ dz\) goes to zero when we substitute z=-y. Is that right?

iocal said:

Oh **** it. Apologies for my french and for like the hundrenth time, thank you.

And well then the final equation \(\displaystyle \displaystyle = -\int_0^{\infty} y\ f(c+y)\ dy + \int_0^{\infty}z\ f(z+c)\ dz\) goes to zero when we substitute z=-y. Is that right?

Click to expand...

"z" and "y" are both dummy variables of integration, so you can replace one by the other in one of the integrands. I have already made one of the integrals negative by reversing the limits.

Good luck learning LaTeX! BTW, the X in the name is the Greek letter Chi, to "tex" is pronounced "tech".

Never mind. I understand your point now. Well like I said thanks a lot man. And I still have a lot to learn so stick around!