Non Homogeneous Poisson Process, failure rates and analysis

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All questions answered but I haven't had a chance to check my work, thanks

Suppose the failure time history of a component is given (where downward -facing arrows indicate failure times):

Assume the failure process follows a NHPP with λ(t) = exp(α+βt). Estimate the model parameters α and β using the MLE.
What is the expected number of failures from the current time until 72 months?
Check if the failure process has a time trend [α = 0.1]. If it does not, find the ROCOF.

(a) A NHPP is defined by a random process which has varying pdfs defined by intensity functions which are a function of time for different time periods. The pdf defining time to first failure (observed over a time period T) for a NHPP is
f_1 (t)=λ(t).e^(-∫_0^T▒〖λ(t^' ).dt^' 〗)
The likelihood function is defined by the product of n elemental pdfs (f_1 (t_i))for failures of first component, thus
L=∏_(i=1)^n▒〖f_1 (t_i ) 〗

Making substitutions λ(t)=e^(α+βt)=e^α e^βt
∫_0^T▒〖λ(t^' ).dt^' 〗=e^α ∫_0^T▒〖e^(βt^' ).dt^' 〗=e^α/β (e^βT-1)
∏_(i=1)^n▒〖e^α e^(βt_i ).e^(-e^α/β(e^βT-1)) 〗
L=∏_(i=1)^n▒〖e^α e^(βt_i ).e^(e^α/β(1-e^βT)) 〗
L=exp{nα+β∑_(i=1)^n▒〖t_i+exp⁡(α)/β[1-exp⁡(βT]〗}

Taking natural logarithms to simplify
Λ=Ln(L)=nα+β∑_(i=1)^n▒〖t_i+exp⁡(α)/β[1-exp⁡(βT)]〗
To find MLE for parameter α we find dΛ/dα and set this to zero
dΛ/dα=n+((1-e^βT))/β e^α=0
Thus
e^α=nβ/(e^βT-1) [1]

To find MLE for parameter β we find dΛ/dβ and set this to zero
dΛ/dβ=∑_(i=1)^n▒〖t_i-e^α/β^2 (1-e^βT ) 〗-(Te^α)/β e^βT=0
Thus
∑_(i=1)^n▒t_i =e^α/β^2 (1-e^βT )+ (Te^α)/β e^βT
Substituting the previous result e^α=nβ/(e^βT-1) in the above
∑_(i=1)^n▒t_i =nβ/(β^2 (e^βT-1) ) (1-e^βT )+nTβ/β(e^βT-1) e^βT
∑_(i=1)^n▒t_i =(nTe^βT)/(e^βT-1)-n/β
Dividing the first term (in the RHS of the above expression) both top and bottom by e^βT which is the equivalent of multiplying by 1 we obtain
∑_(i=1)^n▒t_i =(nTe^βT/e^βT)/((e^βT-1)/e^βT )-n/β

∑_(i=1)^n▒t_i =nT/(1-e^(-βT) )-n/β [2]

For a full mathematical treatment see article here
Equation [2] in the above cannot be solved analytically so we choose to solve for MLE of β value numerically. I complied an excel spread sheet and input test values for β that yielded the sum of t_i values
Examining the data we see we have n=7 in the data set,
∑_(i=1)^n▒t_i =5.2+10.8+18.3+28.8+33.3+44.1+49.2=189.7
Therefore I found
MLE β=0.0004115
From this value of β in equation [1] we therefore find the MLE for parameter α
MLE α=(7×0.0004115)/(e^(0.0004115×54)-1)=0.128195
Thus the process is modelled by intensity factor
λ(t)=e^(0.128195+0.0004115t)

(b) For the NHPP process we estimated in (a) that the intensity function is
λ(t)=e^(0.128195+0.0004115t)
The number of failures between 54 and 72 hours is given by the integral
N_72 (t≥54)=∫_54^72▒〖λ(t).dt=∫_54^72▒e^(0.128195+0.0004115t) .dt〗
N_72 (t≥54)=e^0.128195 (_54^72)|e^0.0004115t/0.0004115| =21

(c) The hazard function is given by
h(t)=(f(t))/(R(t))
For NHPP the pdf is given by f(t)=λ(t).e^(-∫_0^T▒〖λ(t).dt〗) and the reliability by

R(t)=e^(-∫_0^T▒〖λ(t).dt〗)
Hence for such a process the hazard function (equal to the intensity function) results
h(t)=(λ(t).e^(-∫_0^T▒〖λ(t).dt〗))/e^(-∫_0^T▒〖λ(t).dt〗) =λ(t)
For the particular case (rounding to one decimal place) α=0.1,β=0.0
h(t)=e^0.1 e^0.0t=e^0.1=1.1052
As β≃0 process has a ...