# What is this equality: $\frac{\theta/(\theta+1)^{x+k}}{\theta^k}=(\frac1{\theta+1})^k(\frac{\theta}{\theta+1})^x$? How to work it out?

I am struggling with this equality, which I found in George C. Canavos Applied Probability and Statistical Methods:

$\frac{\theta / (\theta + 1)^{x+k}}{\theta^k} = \left(\frac{1}{\theta + 1}\right)^k \left(\frac{\theta}{\theta + 1}\right)^x$

How do I work it out? Is it a known equality?

Cheers

Update

In sight of your kind answers, I start feeling this may be a typo in Canavo's book. I attach a capture from the book's page where the equality appears. As you see, there is no outer parenthesis:

• Before calling that a typo one should perhaps view the larger picture. Perhaps Cavano is using the (non-standard) convention that dash-division has higher priority than exponentiation? In which case you don't need a parenthesis... May 10, 2017 at 9:35
• I personally would find very unlikely to have division higher priority than exponentiation, but my domain of expertise is computer science more than maths, so I will observe your opinion. May 10, 2017 at 13:20
• By the way, thank you for your solution and follow-up. May 10, 2017 at 13:24

That's trivial, but you may need a parenthesis more on the LHS:

\begin{align*} {(\theta/(\theta+1))^{x+k}\over \theta^k}&= {\theta^{x+k}\over(\theta+1)^{x+k}}{1\over \theta^k}\\ &= {\theta^{x}\over(\theta+1)^{x}}{\theta^{k}\over(\theta+1)^{k}}{1\over \theta^k}\\ &= {\theta^{x}\over(\theta+1)^{x}}{1\over(\theta+1)^{k}}\\ &=\left({\theta\over\theta+1}\right)^x\left({1\over\theta+1} \right)^k \end{align*}

Note that that interpretation of $\theta/(\theta+1)^{x+k}$ is required for the equality to be correct. Let for example $x=k=1$ and $\theta=1/2$ and you get the RHS to be $2/9$ and with that interpretation you get the same on the LHS, but without the parenthesis you get the LHS to be $4/9$. Note that the extras in the book doesn't help as this is just a factor $1$ in this case.

• It may be the catch, that the original formula is wrong. As you say, it is trivial once fixed adding the LHS parenthesis. May 10, 2017 at 8:40

$$\frac{\theta / (\theta + 1)^{x+k}}{\theta^k} = \left(\frac{1}{\theta + 1}\right)^k \left(\frac{\theta}{\theta + 1}\right)^x$$

$$\frac{\left[\theta / (\theta + 1)\right]^{x+k}}{\theta^k} = \left(\frac{1}{\theta + 1}\right)^k \left(\frac{\theta}{\theta + 1}\right)^x$$

So:

\begin{align} &\frac{\theta^{x+k}}{(\theta+1)^{x+k}\theta^k}\\ &=\frac{\theta^x}{(\theta+1)^{x+k}}\\ &=\frac{\theta^x}{(\theta+1)^x(\theta+1)^k}\\ &=\frac{\theta^x}{(\theta+1)^x(\theta+1)^k}\\ &=\frac{\theta^x}{(\theta+1)^x}\frac{1}{(\theta+1)^k}\\ &=\left(\frac{\theta}{\theta+1}\right)^x\left(\frac{1}{\theta+1}\right)^k\\ &=\left(\frac{1}{\theta+1}\right)^k\left(\frac{\theta}{\theta+1}\right)^x \end{align}

which is true.

I'll assume that $\theta,x,k \in \mathbb{R}$ and $\theta \neq -1$.

Multiply both sides by $\theta^k$ and combine the exponents on the right side:

$\frac{\theta}{\left(\theta+1\right)^{x+k}}=\left(\frac{\theta}{\theta + 1}\right)^{x+k}$

Multiply both sides by $\left(\theta+1\right)^{x+k}$

$\theta=\theta^{x+k}$

So this should hold for $x+k=1$ and/or $\theta=1$. I haven't seen this equality before so I'm unsure if it's well known, but hope this helps.

• Thanks for this, you really sticked to the formula as posted. It may be a typo. I will wait for the experts' opinion. May 10, 2017 at 8:42