Using the formula $$ P(X =x|Y =y) = \frac{ P(X =x)∗\sum{_{x\in D_k(y)}}{P(K =k)} }{\sum{_{k|y \in C(k)}}P(K =k)∗P(X =D_k(y)). } $$ to show is that if all 26 keys $z$ are equally probable, the shift cipher is perfectly secure - for one-letter plain texts.
Characterization of perfect security is given by $:=$ Let P, C, and K be the sets of possible plaintexts, ciphertexts and keys of an encryption method and E a mapping from P to C. Nextsis $|P|=|C|=|K|$ and $P(X =x) >0 $. The encryption method E is perfectly secure if and only if every key is chosen with the same probability and for all x ∈ P and y ∈ C there is exactly one key with $E_k(x) = y$.
This means that there must be as many keys as there are plaintexts. In particular, it follows that the keys must be at least as long as the plain texts, which is obviously problematic with large plain texts.
So I am a bit lost here. Do I need to get a result of 1.0 (100%) in the result if a put some well argument values into the formula? Since we only have one letter in plaintext, as we encrypt these we will still be one possible cypher and all other potential cyphers will have the same probability because it is given that all keys are equally probable? Could anybody please explain how to solve this in a right way?