4
$\begingroup$

Consider a chain which is not Markov that waits a time $T^{*}$ before leaving the current state, where $T^{*}$ has uniform distribution over the set of times $\{1, 2, 3, 4\}$ . I would like to show that it does not hold the memoryless property, i.e: $$\Bbb P(T \gt t+s \mid T \gt t)= \Bbb P(T \gt s)$$

So, here's how I tackle the problem:

I deduce that $T^{*}\sim\mathcal U\{1,4\}$ (discretely uniformly distributed on the interval $[1,4]$. So, the support is $x\in \{1,2,3,4\}$

The cumulative distribution function is:

$$\text{CDF}= \frac{\lfloor x \rfloor - 1+1}{4} = \frac{\lfloor x \rfloor}{4}= \Bbb P(T \leq x)$$

$$\Rightarrow 1- \text{CDF} = \Bbb P(T \geq x)= 1- \frac{\lfloor x \rfloor}{4}= \frac{4-\lfloor x \rfloor}{4}$$

$$\Rightarrow \Bbb P(T \geq s+t | T \geq t)= \frac{\Bbb P(T\geq s+t ; T\geq t)}{\Bbb P(T \geq t)}= \frac{\Bbb P(T \geq s+t)}{\Bbb P(T \geq t)} = \frac{\frac{4- \lfloor s+t \rfloor}{4}}{\frac{4- \lfloor t \rfloor}{4}}= \frac{4 - \lfloor s+t \rfloor}{4 - \lfloor t \rfloor} \neq \Bbb P(T\geq s) = \frac{4- \lfloor s \rfloor}{4}$$ Hence $T^{*}$ does not hold the memoryless property

Is this alright?

Also, if $(W_k)_{k\geq}$ would be a stochastic process constructed so that it stays in each state $i$ for a time distributed to $T^{*}$, what would be an intuitive explanation of why this is not a Markov chain?

$\endgroup$
3
  • $\begingroup$ The process you are describing is not a Markov chain... $\endgroup$ – Math1000 Feb 10 '20 at 20:22
  • $\begingroup$ @Math1000 yes, my bad, it's not $\endgroup$ – user634512 Feb 10 '20 at 20:27
  • 1
    $\begingroup$ Is this $T^*$ the same as $T$? $\endgroup$ – Saad May 30 '20 at 14:55
1
+50
$\begingroup$

I assume your $T$ and $T^*$ are the same thing. The memoryless property is that, for all values of $s,t$:

$$P(T > t + s \mid T > t) = P(T > s)$$

So to show that $T$ lacks the memoryless property, all you need is to find one counter-example - you do not need to show that the equation is invalid generally.

In this case almost every example you find is already a counter-example, e.g. pick $s = t = 2$:

$$P(T > 2 + 2 \mid T > 2) = 0 \neq P(T > 2) = 1/2$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.