# Prove Bernoulli Process is Markov Chain

• Sep 5th 2009, 04:42 PM
Jimmy_W
Prove Bernoulli Process is Markov Chain
Let \$\displaystyle Z_1, Z_2, ...\$ be a Bernoulli Process with success parameter p. Prove if

\$\displaystyle Y_0, Y_1,...\$, where \$\displaystyle Y_{k+1} = max(0, Y_k + 2Z_{k+1} - 1)\$ for \$\displaystyle k = 0,1,...\$ and \$\displaystyle Y_0 = 0\$

and

\$\displaystyle X_0, X_1,..\$ where \$\displaystyle X_{k+1} = Z_{k+1} + Z_k\$ for \$\displaystyle k = 0,1,...,\$ and \$\displaystyle X_0 = 0\$

are Markov chains.
• Sep 8th 2009, 06:40 PM
Jimmy_W
Does anyone have any idea how to go about these?
• Sep 9th 2009, 01:58 AM
pedrosorio
\$\displaystyle X_2 = Z_2 + Z_1\$

Let us suppose \$\displaystyle X_2 = 1\$ then, either \$\displaystyle Z_2=1 \mbox{ and } Z_1=0 \mbox{or} Z_1=0 \mbox{ and } Z_2=1\$ with 0.5 probability of each case.

Given this, and knowing that \$\displaystyle X_3 = Z_3 + Z_2\$

\$\displaystyle P(X_3 = 1|X_2 = 1) = p*0.5 + (1-p)*0.5 = 0.5\$

However, if we know that \$\displaystyle X_1 = Z_1 + Z_0 = 2\$ we know that \$\displaystyle Z_1 = 1\$

Taking this into account, we get that,\$\displaystyle X_1 = 2\$ and \$\displaystyle X_2 = 1\$, imply \$\displaystyle Z_2 = 0\$.

Therefore,

\$\displaystyle P(X_3 = 1 | X_2 = 1, X_1 = 2) = p\$

It doesn't seem to be a Markov Chain