Transfer entropy value between 0 and 1 Given two variables, X and Y, there is a way of obtaining a Mutual Information value between 0 and 1 by:
MI_normalised=MI_original/sqrt(H(X)*H(Y));

where H(X) and H(Y) are entropies of X and Y respectively.
Just wondering if there is a similar operation to obtain a Transfer Entropy value between 0 and 1. If so, what is it?
Any help much appreciated.
 A: Transfer entropy is defined as:
$$T_{X \to Y} = H(Y_t \mid Y_{t-1 : t-L}) - H(Y_t \mid Y_{t-1 : t-L}, X_{t-1 : t-L})$$
where $t$ is the current time and $L$ is the history length.
If $X$ perfectly predicts $Y$ then
$$H(Y_t \mid Y_{t-1 : t-L}, X_{t-1 : t-L}) = 0$$
and $T_{X \to Y}$ therefore has a maximum value of $H(Y_t \mid Y_{t-1 : t-L})$.
If $X$ contains no information about $Y$ then
$$H(Y_t \mid Y_{t-1 : t-L}, X_{t-1 : t-L}) = H(Y_t \mid Y_{t-1 : t-L})$$
and $T_{X \to Y}$ therefore has a minimum value of $0$.
So, if the goal is to normalize transfer entropy to the range $[0, 1]$, it seems reasonable to divide it by its maximum possible value:
$$\frac{T_{X \to Y}}{H(Y_t \mid Y_{t-1 : t-L})}$$
I don't know whether this is standard; just working off the definition. So please check that this makes sense.
A similar definition is given in

Gourevitch and Eggermont (2007). Evaluating Information Transfer
  Between Auditory Cortical Neurons

and other neurophysiology papers (search for 'normalized transfer entropy'). They often include an additional correction term based on shuffled data to correct for bias when estimating from finite, noisy signals.
A: I think that in practice I would use user20160's answer, but I'd like to offer a few warnings and a possible alternative.
Warning 1: The definition $T_{X\rightarrow Y} / H(Y_t | Y_{t-1:t-L})$ only works for discrete random variables. If you you have continuous random variables, then $H(Y_t | Y_{t-1:t-L})$ could be negative and your quantity would no longer be between zero and one. 
Warning 2: Suppose that $H(Y_t | Y_{t-1:t-L}) = \epsilon$ and $H(Y_t | Y_{t-1:t-L},X_{t-1:t-L}) = 0$. You will get a score of 1. In the limit of $\epsilon\rightarrow 0$, you continue to have a score of 1 even though, in some sense, the amount of information transferred is going to zero. 
Alternative 1: If you are using discrete random variables, note that $0 \leq T_{X\rightarrow Y} \leq H(Y) \leq \log k$, where $k$ is the number of discrete possibilities $Y$ can take. Then if you define $T_{X\rightarrow Y} / 
log k$, you still get something between zero and one, but it will only be one if $Y$ is maximally uncertain and $X$ is a perfect predictor. This would bypass warning 2, if that is a concern in your case. 
Alternative 2: For the continuous case you could define your score as $T_{X\rightarrow Y} / I(Y_t ; Y_{t-1:t-L},X_{t-1:t-L})$. The denominator is always non-negative and $T_{X\rightarrow Y} \leq I(Y_t ; Y_{t-1:t-L},X_{t-1:t-L})$ (you could see that using the chain rule, e.g.). You would get one if $I(Y_t ; Y_{t-1:t-L}) = 0$ which could loosely be interpreted as saying all the information that can be predicted about $Y_t$ comes from $X$, and not from $Y$'s past. 
