Expected entropy in the output of a random oracle
The expected entropy in the output of a $h$-bit random oracle fed with random $h$-bit input is close to $h-0.8272$ bit, for even moderate $h$ (e.g. at least $32$). As $h$ grows, that expected entropy becomes arbitrary close to $h-\eta$ bit with
$$\begin{align}\eta&=\frac 1{e\ln(2)}\sum_{i=1}^\infty\frac{\ln(i+1)}{i!}\\
&=0.82724538915300508343173\dots\text{bit}\end{align}$$
where the sum is given by A193424.
Proof, where I'll be using $a\approx b$ as a convenient shorthand for $\displaystyle\lim_{h\to\infty}\frac a b\ =\ 1$
- For a particular distribution implemented by the oracle, let $n_j$ be the number of output values appearing exactly $j$ times among the outputs for all inputs. The exact entropy $H$ for that particular distribution can be computed from the $n_j$ by applying the definition of entropy, giving
$$\begin{align}H&=\sum_{j=1}^{2^h}n_j\;\frac j {2^h}\;\log_2\left(\frac{2^h}j\right)\\
&=\frac h{2^h}\sum_{j=1}^{2^h}j\;n_j\;-\frac 1{2^h}\sum_{j=1}^{2^h}n_j\;j\;\log_2(j)\end{align}$$
where we have (by merely counting what all inputs lead to)
$$\sum_{j=1}^{2^h}j\;n_j\;=2^h$$
thus
$$h-H=\frac 1{2^h}\sum_{j=1}^{2^h}n_j\;j\;\log_2(j)$$
- For fixed $j$ and as $h$ grows, by counting of the possibilities, we can establish that for random distribution, odds that any particular value is reached $j$ times is $\displaystyle\approx\frac 1{e\;j!}$. Thus for fixed $j$ and as $h$ grows, the expected $n_j$ is $\displaystyle\approx\frac{2^h}{e\;j!}$.
- In the exact expression of $h-H$, all the terms in the sum are non-negative. To obtain an asymptotic of the expected $h-H$ when $h$ grows, we can thus replace $n_j$ by its expected value, and obtain that when $h$ grows the expected value of $h-H$ is $\displaystyle\approx\frac 1 e\sum_{j=1}^\infty\frac{j\;\log_2(j)}{j!}$.
- The stated result follows by defining $i=j-1$, and removing the sum's first term, which is zero.
I've been unable to locate a source; the closest I found is an empirical derivation of $\eta$ to 4 decimals by Andrea Röck: Collision Attacks based on the Entropy Loss caused by Random Functions, WEWoRC 2007, slides; with more in her thesis.
My first empirical derivation was using a program which draws $2^h$ pseudo-random $h$-bit values and counts how many values are reached how many times; for $h=35$ (the largest I could do with 20GB RAM), three runs gave:
run 1 run 2 run 3
0 12640123427 36.79% 12640183855 36.79% 12640308584 36.79%
1 12640408212 36.79% 12640365800 36.79% 12640104651 36.79%
2 6320124091 18.39% 6320013534 18.39% 6320174710 18.39%
3 2106681541 6.13% 2106762262 6.13% 2106726749 6.13%
4 526645276 1.53% 526674914 1.53% 526679947 1.53%
5 105334156 0.31% 105325000 0.31% 105330269 0.31%
6 17561277 0.05% 17551924 0.05% 17556150 0.05%
7 2507918 0.01% 2508727 0.01% 2505282 0.01%
8 313971 0.00% 313943 0.00% 313406 0.00%
9 34748 0.00% 34553 0.00% 34755 0.00%
10 3424 0.00% 3542 0.00% 3546 0.00%
11 291 0.00% 287 0.00% 292 0.00%
12 31 0.00% 24 0.00% 24 0.00%
13 4 0.00% 3 0.00% 2 0.00%
14 1 0.00% 0 0.00% 1 0.00%
15+ 0 0.00% 0 0.00% 0 0.00%
entropy 34.172763 bit 34.172758 bit 34.172751 bit
Application to the question: the entropy for the output of SHA-256 truncated to its first $128$ bits when fed a random $128$-bit input is about $127.173$ bit, down from very close to $128$ bit before truncation (see final note). The truncation does not halve the entropy, because the halves are not independent. The right line of thought is that SHA-256 truncated to its first $128$ bits is a fine $128$-bit hash, and behaves like a random oracle.
Note: if we consider a random function from $\{0,1\}^{128}$ to $\{0,1\}^{256}$, most likely there is a small $k$ (most often $0$ or $1$, sometime $2$, rarely $3$ or more) such that $k$ outputs have exactly two corresponding inputs, $2^{128}-2k$ outputs have exactly one, and $2^{256}-2^{128}+2k$ outputs have none (odds that any output is reached three or more times are negligible).
Therefore, in this most likely case, the entropy on output of that function when fed a random $128$-bit input is $-k\;2^{-127}\log_2(2^{-127})-(2^{128}-2k)\;2^{-128}\log_2(2^{-128})$, that is $128-k\;2^{-127}$.
The best model we have for SHA-256 (not truncated) for $128$-bit input is a particular function chosen at random among functions from $\{0,1\}^{128}$ to $\{0,1\}^{256}$, thus we can conclude the entropy for the output of SHA-256 when fed a random $128$-bit input is likely exactly $128-k\;2^{-127}$ for $k\in\{0,1,2\}$, which is very nearly $128$ bit, down to about the 37th decimal places.
