# Proof of lack of memoryless property

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?

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

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