Calculating E[x] for f(x)=e^-2|x| distribution in the reals (x e R)

[HappyN]

I want to calculate E[x] of the following continuous distribution having density: f(x)=e^-2|x|
for x in the reals (x e R)

I did the calculation with integral bounds infinity and minus infinity, are these the right bounds to use since we are only told x e R?
I got 0 as the answer, can someone tell me if they get the same?

 

[tiny-tim]

Hi HappyN!

I'm guessing that you didn't adjust the integral for negative x to take account of the |x|

(it usually makes the integral negative if you forget)

 

[HappyN]

do you mean my bounds are wrong?
i'm not quite sure of what you mean by adjusting the integral for negative x?

 

[tiny-tim]

Show us your full calculations.

 

[HappyN]

To calculate E[X] I did: ∫xf(x) dx (integral bounds between minus ∞ and ∞ - sorry don't know how to type it properly!)
using integration by parts, i got:
E[x]=[-x/2 e^-2|x|] + [1/4 e^-2|x|] (bounds evaluated between -∞ and ∞)
=(-∞/2 e^-2|∞|) - (∞/2 e^-2|∞|) + (-1/4 e^-2|∞| + 1/4 e^-2|∞|)
=-∞e^-2|∞|
which is 0?
therefore E[x]=0?

 

[HallsofIvy]

It should be obvious by symmetry that E[x]= 0.

 

[tiny-tim]

Hi HappyN!

(just got up :zzz: …)

E[x]=[-x/2 e^-2|x|] + [1/4 e^-2|x|] (bounds evaluated between -∞ and ∞)

(try using the X² icon just above the Reply box )

if x < 0, then eg d/dx e^-2|x| = d|x|/dx d/d|x| e^-2|x|

= (-1) -2e^-2|x| …

the d|x|/dx makes everything negative for negative x !