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Just for the sake of providing some context, I'm dealing with the following:

Alice's smartphone plan includes a voicemail service with a maximum storage capacity of two voice messages.

On a daily basis, Alice listens and responds to voice messages based on her available time. Every morning, Alice can respond to a maximum of 0 messages with probability $p_0=0.2$, 1 messages with probability $p_1=0.3$, and 2 messages with probability $p_2=0.5$. The message is erased from the voice mailbox once Alice answers it.

Alice receives the day before 0 messages with probability $q_0=0.1$, 1 messages with probability $q_1=0.5$, and 2 messages with probability $q_2=0.4$. All new messages will be discarded if the voice mailbox is already full. Assume that no voice messages are received in the mornings during the time she checks her phone.

The corresponding Markov chain model is as seen in the figure below:

discrete-time homogeneous Markov chain

There are three states $V=\{0,1,2\}$ in the Markov chain above, showing the amount of messages left in Alice's voicemail box after she checks her voicemails in the mornings. We determine that $p_{00}=0.7$.

The problem now asks to compute the transition probabilities $p_{01}$, $p_{11}$, and $p_{22}$.

This is what I came up with:

$$p_{01}=(q_1*p_0)+(q_2*p_1)=(0.5*0.2)+(0.4*0.3)=0.1+0.12=0.22$$ $$p_{11}=(q_1*p_0)=(0.5*0.2)=0.1$$ $$p_{22}=(q_2*p_0)=(0.4*0.2)=0.08$$

This seems about right.

Could someone tell me if my calculations are correct? Thanks

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1 Answer 1

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I agree with your answers for $\ p_{00}\ $ and $\ p_{01}\ $ but not with those for $\ p_{11}\ $ or $\ p_{22}\ $.

If, for instance, there are two messages in Alice's voicemail box after she has checked her voicemails on one day, then there will still be those two messages in it when she comes to check her voicemail box on the next, regardless of how many more voicemails (if any) have arrived (and necessarily been discarded) in the meantime. There will still be two voicemails left in the voicemail box after Alice has finished dealing with it if and only if the maximum number she can respond to is $0$, which has probability $\ p_0\ $. Thus, $\ p_{22}=p_0\ $.

Here are the values I get for all $\ 9\ $ of the transition probabilities: \begin{align} p_{00}&=q_0+q_1(p_1+p_2)+q_2p_2=0.7\\ p_{01}&=q_1p_0+q_2p_1=0.22\\ p_{02}&=q_2p_0=0.08\\ \\ p_{10}&=q_0(p_1+p_2)+(q_1+q_2)p_2=0.53\\ p_{11}&=q_0p_0+(q_1+q_2)p_1=0.29\\ p_{12}&=(q_1+q_2)p_0=0.18\\ \\ p_{20}&=p_2=0.5\\ p_{21}&=p_1=0.3\\ p_{22}&=p_0=0.2\ . \end{align}

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  • $\begingroup$ Thank you very for you answer @lonzaleggiera it makes perfect sense. My approach for the transitions was $$p_{11}=P("1 \space messages \space received \space and \space not \space answered" )=q_1p_0$$ and $$p_{22}=P("2 \space messages \space received \space and \space both \space not \space answered")=q_2p_0$$ $\endgroup$
    – nimen55290
    May 15, 2022 at 17:55
  • $\begingroup$ Btw, I posted another question here which is based on the same markov chain but with conditional probabilities. Do you mind having a look at it as well? I would really appreciate some feedback @lonzaleggiera $\endgroup$
    – nimen55290
    May 15, 2022 at 17:59

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