Solution:
M(t)=12(2+et+6e3t+3e6t)eβtβ
=12(2et+e2t+6e4t+3e7t)1β=121β(2et+e2t+6e4t+3e7t)β1
On differentiating w.r.t t,
Mβ²(t)=12β1β(2et+e2t+6e4t+3e7t)β2(2et+2e2t+24e4t+21e7t)
=β12(3e7t+6e4t+e2t+2et)221e7t+24e4t+2e2t+2etβ
Again, differentiating,
Mβ²β²(t)=12(3e6t+6e3t+et+2)3eβt(441e12t+846e9t+9e7t+444e6t+72e4tβ12e3t+4e2t+6et+4)β
Now, Mβ²(0)=β12(3e0+6e0+e0+2e0)221e0+24e0+2e0+2e0β=β172849β
And Mβ²β²(0)=12(3+6+1+2)3441+846+9+444+72β12+4+6+4β=10368907β
Then, variance=Mβ²β²(0)β[Mβ²(0)]2=10368907ββ(β172849β)2
So, variance=0.08667
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