Dash and dot probabilities

Question

This is problem 1.41(b) in Casella and Berger's Statistical Inference.

Consider telegraph signals "dot" and "dash" sent such that $$\mathbb{P}(\text{dot sent}) = \dfrac{3}{7}$$ and $$\mathbb{P}(\text{dash sent}) = \dfrac{4}{7}$$ where erratic transmissions cause a dot to become a dash with probability $\dfrac{1}{4}$ and a dash to become a dot with probability $\dfrac{1}{3}$.

It isn't too hard to see, using Bayes' Theorem, that $$\mathbb{P}(\text{dash sent} \mid \text{dash received}) = \dfrac{32}{41}$$ and then it states:

Assuming independence between signals, if the message dot-dot was received, what is the probability distribution of the four possible message that could have been sent?

There are four possible messages:

1. dot-dot
2. dot-dash
3. dash-dot
4. dash-dash.

[My work here originally was completely wrong, and has been removed.]

Answer 1

Let's take it from your last equation, where you assume conditional independence:

$$\mathbb{P}(\text{dot-dot received} \mid \text{dot-dot sent}) = \left[\mathbb{P}(\text{dot received} \mid \text{dot-dot sent}) \right]^2.$$

To make things clearer, we will number the signals:

$$\mathbb{P}(\text{dot1-dot2 received} \mid \text{dot1-dot2 sent}) = \mathbb{P}(\text{dot1 received} \mid \text{dot1-dot2 sent})\cdot \mathbb{P}(\text{dot2 received} \mid \text{dot1-dot2 sent}).$$

Now, you're being told that the signals are independent, and therefore knowing that the second signal sent was a dot doesn't say anything about the first signal you received (and vice-versa). This means that you get the equality you wanted:

$$\mathbb{P}(\text{dot1-dot2 received} \mid \text{dot1-dot2 sent}) = \mathbb{P}(\text{dot1 received} \mid \text{dot1 sent})\cdot \mathbb{P}(\text{dot2 received} \mid \text{dot2 sent}).$$

Or written differently: $$\mathbb{P}(\text{dot-dot received} \mid \text{dot-dot sent}) = \left[\mathbb{P}(\text{dot received} \mid \text{dot sent}) \right]^2.$$

Answer 2

If you consider this in the context of the four outcomes, and given the independence of signals, it's perhaps easier to understand.

Certainly independence means that the signal received can only depend on the signal sent, not on the next or previous signal. It also means - as I read it - that the signal sent also does not depend on preceding or following signals. This is perhaps where you hit a wall that I don't see.

Overall we have $\frac{27}{84}$ probability of receiving an "honest" dot, and $\frac{16}{84}$ of a false dot. Then p(dot sent $\mid$ dot rec'd) $= \frac{27}{43}$ and p(dash sent $\mid$ dot rec'd) $= \frac{16}{43}$, and due to independence the four possible original messages for a received dot-dot just multiply these up.

• p(dot-dot sent $\mid$ dot-dot rec'd) $= \frac{27}{43}\cdot\frac{27}{43}$
• p(dot-dash sent $\mid$ dot-dot rec'd) $= \frac{27}{43}\cdot\frac{16}{43}$
• p(dash-dot sent $\mid$ dot-dot rec'd) $= \frac{16}{43}\cdot\frac{27}{43}$
• p(dash-dash sent $\mid$ dot-dot rec'd) $= \frac{16}{43}\cdot\frac{16}{43}$