provides a general construction) but up to now there did not exist better bounds. For $(2,1,1)$, we have already seen that the minimal possible field size is $3$, i.e. much smaller than all these bounds.

Probability
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We want to compute the probability that a non-catastrophic convolutional code with $(n-k)\mid\delta$ and generic row degrees $\nu=\frac{\delta}{n-k}$ is complete MDP. Therefore, we assign again to each code the unique parity-check matrix from Remark \[kh\]. This is possible since permutation of the columns of the parity-check matrix does not influence the property to be complete MDP. With these definitions/settings, one gets the following theorem:

 \
If $|\mathbb F|>\min\{N_1,N_2\}$, the probability for a complete MDP convolutional code is at least $$\max\left\{1-\frac{(L+1)(n-k)\binom{(L+1+\frac{\delta}{n-k})n}{(L+1)(n-k)}\cdot t}{1-t^k+O(t^{k+1})},1-\frac{(L+1)(n-k)(\frac{\delta n}{n-k} +k+1)^{(n-k)(L+1)}\cdot t}{1-t^k+O(t^{k+1})} \right\}$$

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The proof is completely analogue to the proof for the probability of MDP convolutional codes.

Sufficient field size for MDP convolutional codes with $\delta<\max\{k,n-k\}$
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In this section, we show a better bound on the necessary field size for MDP convolutional codes for the case that $\delta<\max\{k,n-k\}$. Because of duality arguments, we mainly have just to solve the case $\delta<k$.

The case $\delta<k$
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To derive an upper bound for the required field size, one could assume 