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Question: convolution boundaries

Question:convolution boundaries

JoyDivisionMan 95 Maple 2025
integration convolution statistics
2 hours ago
0

Greetings, Friends.

I have a function f(t) which is a density function. It has a [0,2] interval and integrates to unity. I convolve f(t) with itself to get g(t). It has a [0,4] interval and integrates to unity. All is well with that and I understand.

BUT, the boundary of g(t) shouild be on the interval [0,4/sqrt(3)]. This is based upon both my simulation and other analyses.

At present, the area of the g(t) function on the [4/sqrt(3), 4] interval accounts for 1.408701% of the area under the curve. Not a "trivial" nor "throwaway" amount.

Does anyone know of a way to "re-scale" or modify my convolved result g(t) so that all of the resultant area under the curve is on the [0,4/sqrt(3)] interval?

My work is attached.

> 

restart
> 

with(Statistics)
> 

f := proc (t) options operator, arrow; piecewise(t <= 0, 0, t <= 2, 6/5-(3/2)*t^2+(3/4)*t^3-(3/80)*t^5, 2 <= t, 0) end proc

proc (t) options operator, arrow; piecewise(t <= 0, 0, t <= 2, 6/5-(3/2)*t^2+(3/4)*t^3-(3/80)*t^5, 2 <= t, 0) end proc

 (1)
> 

plot(f(t), t = 0 .. 2)

 
> 

int(f(t), t = 0 .. 2)

1

 (2)
> 

simplify(int(f(tau)*f(t-tau), tau = -infinity .. infinity))

(1/1971200)*piecewise(t <= 0, 0, t <= 2, t^11-220*t^9+1320*t^8+7920*t^7-103488*t^6+147840*t^5+887040*t^4-2365440*t^3+2838528*t, t <= 4, -(t^4+28*t^3+228*t^2+536*t+80)*(t-4)^7, 4 < t, 0)

 (3)
> 

g := unapply(simplify(int(f(tau)*f(t-tau), tau = -infinity .. infinity)), t)

proc (t) options operator, arrow; (1/1971200)*piecewise(t <= 0, 0, t <= 2, t^11-220*t^9+1320*t^8+7920*t^7-103488*t^6+147840*t^5+887040*t^4-2365440*t^3+2838528*t, t <= 4, -(t^4+28*t^3+228*t^2+536*t+80)*(t-4)^7, 4 < t, 0) end proc

 (4)
> 

plot(g(t), t = 0 .. 4)

 
> 

int(g(t), t = 0 .. 4)

1

 (5)
> 

evalf(int(g(t), t = 0 .. 4/sqrt(3)))

.98591299

 (6)
> 

evalf(int(g(t), t = 4/sqrt(3) .. 4))

0.1408701e-1

 (7)
 

NULL

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