Algebraic problem for satisfying a given equation

Question

I'm trying to solve the following exercise:

My backward equation looks like: $P_{i,j}'(t) = i\lambda P_{i+1,j}(t) - i\lambda P_{i,j}(t) $

So i started with differentiating $P_{i,j}(t)$:

${j-1 \choose i-1}(e^{-i\lambda t}(1-e^{-\lambda t})^{j-i})'$ By using $(uv)' = u'v + uv'$

${j-1 \choose i-1}-i\lambda (e^{-i\lambda t}(1-e^{-\lambda t})^{j-i}) + {j-1 \choose i-1}(j-i)\lambda (e^{-(i+1)\lambda t}(1-e^{-\lambda t})^{j-i-1}) $

where the first term is cleary $-i\lambda P_{i,j}(t) $.

I don't see how i can turn $ {j-1 \choose i-1}(j-i)$ into ${j-1 \choose i}i$ so the second term equals $i\lambda P_{i+1,j}(t) $

Answer 1

You seem to have the wrong definition of $P_{i,j}$ it should be ${j-1 \choose i-1}$, not ${j-i \choose i-1}$. If you do the same thing with this corrected version, then $$ \begin{align*} {j-1 \choose i-1}(j-i) &= \frac{(j-1)!}{(i-1)!(j-i)!}(j-i) \\ &= \frac{(j-1)!}{(i-1)!(j-i-1)!} = \frac{(j-1)!}{((i+1)-1)!(j-(i+1))!}i \\ &= {j-1 \choose (i+1)-1}i \end{align*} $$

Answer 2

$$P_{ij}(t)=\binom{j-1}{i-1}e^{-i\lambda t}\left(1-e^{-\lambda t}\right)^{j-i}\\\begin{align*}P'_{ij}(t)&=\binom{j-1}{i-1}\left(-i\lambda e^{-i\lambda t}(1-e^{-\lambda t})^{j-i}+\lambda(j-i)e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{j-i-1}\right)\\&=-i\lambda P_{ij}(t)+\lambda(j-i)\binom{j-1}{i-1}e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{j-(i+1)}\\&=-i\lambda P_{ij}(t)+\lambda\binom{j-1}{(i+1)-1}e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{j-(i+1)}\\&=-i\lambda P_{ij}(t)+\lambda P_{i+1,j}(t)\end{align*}$$... which follows from the fact that $\binom{j-1}{(i+1)-1}=\binom{j-1}i=(j-1)(j-2)\cdots(j-(i-1))(j-i)$