@@ -130,10 +130,10 @@ Example for *n*\ =4:
130130:math: `x_0 ` :math: `x_1 ` :math: `x_2 ` :math: `x_3 ` Probability of Pattern Probability of +1 byte
131131============== ============== ============== ============== ====================== ======================
132132any any any :math: `=0 ` |fp0 | :math: `1 `
133- any any :math: `=0 ` :math: `\ne 0 ` |fp1 | :math: `P(x_3 \le 1 | x_3 \ne 0 )`
134- any :math: `=0 ` :math: `\ne 0 ` :math: `\ne 0 ` |fp2 | :math: `P(x_3 \le 2 | x_3 \ne 0 )`
135- :math: `=0 ` :math: `\ne 0 ` :math: `\ne 0 ` :math: `\ne 0 ` |fp3 | :math: `P(x_3 \le 3 | x_3 \ne 0 )`
136- :math: `\ne 0 ` :math: `\ne 0 ` :math: `\ne 0 ` :math: `\ne 0 ` |fp4 | :math: `P(x_3 \le 4 | x_3 \ne 0 )`
133+ any any :math: `=0 ` :math: `\ne 0 ` |fp1 | :math: `P(x_3 \le 1 \vert x_3 \ne 0 )`
134+ any :math: `=0 ` :math: `\ne 0 ` :math: `\ne 0 ` |fp2 | :math: `P(x_3 \le 2 \vert x_3 \ne 0 )`
135+ :math: `=0 ` :math: `\ne 0 ` :math: `\ne 0 ` :math: `\ne 0 ` |fp3 | :math: `P(x_3 \le 3 \vert x_3 \ne 0 )`
136+ :math: `\ne 0 ` :math: `\ne 0 ` :math: `\ne 0 ` :math: `\ne 0 ` |fp4 | :math: `P(x_3 \le 4 \vert x_3 \ne 0 )`
137137============== ============== ============== ============== ====================== ======================
138138
139139.. |fp0 | replace :: :math: `P(x_3 =0 )`
@@ -146,7 +146,7 @@ Multiply the last two columns, and sum for all rows. For a message of length
146146*n * where :math: `1 \le n \le 254 `, the general equation for the
147147probability of the +1 byte is:
148148
149- .. math :: P(x_{n-1} \le n| x_{n-1}\ne 0) \prod_{k=0}^{n-1} P(x_k\ne 0) + \sum_{i=0}^{n-2} \left[ P(x_{n-1} \le (n-1-i)| x_{n-1}\ne 0) P(x_i=0) \prod_{k=i+1}^{n-1} P(x_k\ne 0) \right] + P(x_{n-1}=0)
149+ .. math :: P(x_{n-1} \le n \vert x_{n-1}\ne 0) \prod_{k=0}^{n-1} P(x_k\ne 0) + \sum_{i=0}^{n-2} \left[ P\bigl (x_{n-1} \le (n-1-i) \vert x_{n-1}\ne 0 \bigr ) P(x_i=0) \prod_{k=i+1}^{n-1} P(x_k\ne 0) \right] + P(x_{n-1}=0)
150150
151151
152152Even Byte Distribution Case
@@ -155,7 +155,7 @@ Even Byte Distribution Case
155155We can simplify this for the simpler case of messages with byte value
156156probabilities that are evenly distributed. In this case:
157157
158- .. math :: P(x_{n-1} \le n| x_{n-1}\ne 0) = \frac{n}{255}
158+ .. math :: P(x_{n-1} \le n \vert x_{n-1}\ne 0) = \frac{n}{255}
159159
160160.. math :: P(x_i\ne 0) = \frac{255}{256}
161161
0 commit comments