Reputation
563
Top tag
Next privilege 1,000 Rep.
Create new tags
Badges
2 13
Newest
 Yearling
Impact
~18k people reached

  • 0 posts edited
  • 0 helpful flags
  • 26 votes cast
Jun
9
revised Curve of intersection of two surfaces
added 2208 characters in body
Jun
6
asked Curve of intersection of two surfaces
Apr
12
asked Integral of $x^{12}(1-x)^8$
Jan
5
awarded  Yearling
Oct
2
awarded  Popular Question
Sep
19
awarded  Popular Question
Jul
29
asked Transform square region to triangular region
Jul
2
awarded  Curious
Jun
20
comment Existence of kth moment
I finally figured out your answer, another great answer, but you are right, a huge overestimate. Here is some Mathematica code comparing the actual integral and your bound for $k=6$. In[3]:= k = 6 Out[3]= 6 In[7]:= Integrate[ Abs[x]^k*Exp[-(x - 3)^2], {x, -[Infinity], [Infinity]}] // N Out[7]= 2551.67 In[8]:= Integrate[Abs[x + 3]^k, {x, -1, 1}] + 2^(2*k + 1)*(k + 1)^k // N Out[8]= 9.63783*10^8
Jun
19
awarded  Informed
Jun
19
asked $E[U^2]=0$ implies $Pr(U=0)=1$
Jun
19
comment Existence of kth moment
$|-2|>1$, but $|-2+3|$ is not less than $4(-2)$.
Jun
19
comment Existence of kth moment
And that's only to the right of $x=1$.
Jun
19
comment Existence of kth moment
Not sure why the second integral is finite. Here is what I got in Mathematica with $k=20$. Integrate[Abs[x + 3]^k*(k + 1)^(k + 1)/x^(2 (k + 2)), {x, 1, [Infinity]}] // N 1.69501*10^38
Jun
19
comment $X<Y$ implies $E[X]<E[Y]$?
Nice replies everyone, very, very helpful.
Jun
18
asked $X<Y$ implies $E[X]<E[Y]$?
Jun
18
comment $X_1$, $X_2$ independent implies $e^{tX_1}$, $e^{tX_2}$ independent
Absolutely wonderful. All of these answers are useful. Haven't got to the Jacobian yet, but it lies here to help me when I get there. Thanks for the help.
Jun
18
accepted $X_1$, $X_2$ independent implies $e^{tX_1}$, $e^{tX_2}$ independent
Jun
18
asked $X_1$, $X_2$ independent implies $e^{tX_1}$, $e^{tX_2}$ independent
Jun
18
accepted Existence of kth moment