# Moments of Transmuted Lindley Distribution

I've been having difficulty in obtaining the exact solution given by the author. I need help as my solution doesn't factorize completely to the same term given by the author.

Obtaining the kth moments entails computing the integral given in the picture. I carried out the integration but I did not get the exact term the author got. I was of the opinion that I may be wrong in my factorization, hence I seek other ideas to obtain the exact solution the author got. Otherwise, I want to be justified that the author is probably wrong.  • Please use mark-up to show your workings so far. – Nick Cox Aug 22 '16 at 20:58
• If you can drop email I could send you a pdf file of my workings. Trying to find my way round the mark-up ish though. Thanks. – Ian Aug 23 '16 at 10:07
• I am suggesting what you should do to improve the public question, not volunteering private help. As it stands, this is too difficult to follow. – Nick Cox Aug 23 '16 at 10:11
• MathJax basic tutorial and quick reference – Glen_b Aug 23 '16 at 12:04
• Please type your question as text, do not just post a photograph (see here). When you retype the question, add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. – gung Aug 23 '16 at 13:33

Focus on the form of the integrand that matches what you can compute. Specifically, the expectation can be written

\eqalign{ \mathbb{E}(X^k) &= A\int_0^\infty x^k(1+x)e^{-\theta x}\left(B + (C+Dx)e^{-\theta x}\right)\mathrm{d}x \\ &=A\int_0^\infty\left((Bx^k + Bx^{k+1})e^{-\theta x} + (Cx^k + (C+D)x^{k+1} + Dx^{k+2})e^{-2\theta x}\right)\mathrm{d}x }

with $A = \theta^2/(1+\theta)$, $B=1-\lambda$, $C=2\lambda$, and $D=2\lambda\theta/(1+\theta)$.

This splits into a sum of five gamma integrals, so we can read off the answer directly: it is $$A\left(Bk!\theta^{-k-1} + B(k+1)!\theta^{-k-2} + Ck!(2\theta)^{-k-1} + (C+D)(k+1)!(2\theta)^{-k-2} + D(k+2)!(2\theta)^{-k-3}\right).$$

Factor out $\theta^{-k-2}k!$ to obtain

$$\frac{Ak!}{\theta^{k+2}}\left(B(\theta+k+1) + C2^{-k-1}\theta + (C+D)(k+1)2^{-k-2} + D(k+2)(k+1)2^{-k-3}\theta^{-1}\right).$$

The term involving $B$ equals

$$\frac{k!}{\theta^k(1+\theta)}(1-\lambda)(\theta+k+1),$$

recognizable in the first part of the answer. The other terms equal

$$\frac{k!\lambda}{\theta^k(1+\theta)^2 2^{k+2}}\left(k^2 + (4\theta+5)k + 4(1+\theta)^2\right).$$

That's not remotely like what the author obtained. Although I, too, may have made some algebraic mistake, it is obvious that some multiple of $k^2$ must be involved, because the original integral includes a multiple of $x^{k+2}e^{-2\theta x}$, which will introduce $(k+2)!=(k^2+3k+2)k!$ upon integration: that $k^2$ cannot be canceled off by any of the other terms.

• Very nice and helpful. I think I have gotten something like this also. Let me try it out to see if the subsequent moments tally with what he got. Thanks a whole bunch @whuber. Please what application did you use in typing these math objects because I seem not to be able to do directly from my keyboard? – Ian Aug 25 '16 at 19:33
• I typed $\TeX$ markup directly. See the help for "MathJax" when you are editing a question or answer. – whuber Aug 25 '16 at 22:09
• From your result, $$E(x)=\frac{1}{\theta(1+\theta)}[(1-\lambda)(\theta+2)+\frac{\lambda}{\theta(1+\theta)2^3}(1+(4\theta+5)+4(1+\theta)^2)$$ which is different from what he got. I believe the author is wrong sort of. – Ian Aug 27 '16 at 5:22