Laplace Transforms — Question 9

Question 9 – Solution

Strategy. Normalization fixes one parameter, while derivatives of the transform at zero recover moments. A density’s peak is not its mean.

Step 1: Normalize and transform. Integration by parts gives $\int_0^\infty te^{-at}\,dt=1/a^2$. Thus $c=a^2$, and exponential shifting yields $$\boxed{F(s)=\frac{a^2}{(s+a)^2},\qquad s>-a.}$$ The nonnegative polynomial-exponential tail diverges at or below $-a$. In particular, $F(0)=1$, as normalization requires.

Step 2: Extract moments and parameters. Since zero lies strictly inside the convergence interval, exponential decay justifies differentiation near zero: $$-F'(0)=\int_0^\infty tf(t)\,dt,\qquad F''(0)=\int_0^\infty t^2f(t)\,dt.$$ The rational expression gives $F'(s)=-2a^2/(s+a)^3$ and $F''(s)=6a^2/(s+a)^4$. Hence $$\mu=\frac 2a,\qquad \sigma^2=\frac 6{a^2}-\frac 4{a^2}=\frac 2{a^2}.$$ From $\mu=4$, we obtain $$\boxed{a=\tfrac 12,\quad c=\tfrac 14,\quad \sigma^2=8,\qquad F(s)=\frac 1{(2s+1)^2}.}$$

Step 3: Test consistency rather than refitting. Eliminating $a$ gives $\boxed{\sigma^2=\mu^2/2}$. Thus mean four forces variance eight, and the pair $(4,4)$ is incompatible with this family. Changing $c$ cannot repair it because normalization already fixes $c=a^2$.

Step 4: Locate the peak. For general $a$, $f'(t)=a^2e^{-at}(1-at)$, positive before $t=1/a$ and negative afterward. The unique density maximum is therefore at $t=1/a=2$, while the mean is four. The long right tail contributes to the mean; the point of largest density measures a different feature. The graph marks both the mode and the mean for $f(t)=te^{-t/2}/4$.