Geometric distribution - memoryless property

[Simon C]

The geometric distribution is said to have the memoryless property P(X > x + n|X > n) = P(X > x).

The proof for the Type 1 Geometric distribution is shown in the ActEd notes Chapter 4 page 7. I assumed this could also be proved for the Type 2 distribution but, on attempting it, I get P(X > x + n|X > n) = P(X > x – 1). For example, under the Type 2 basis I calculate that P(X > 2 + 4|X > 4) = P(X > 1).

I am fairly sure that my algebra is correct here, especially as the Type 2 distribution is "one out" from the Type 1 distribution. However I am struggling to interpret the meaning of this result or understand why P(X > x + n|X > n) = P(X > x) does not still apply in the Type 2 situation.

I calculate P(X <= x) to be 1 - q^x for the Type 1 distribution and 1 - q^(x + 1) for the Type 2 distribution. This gives P(X > x) as q^x for the Type 1 distribution and q^(x + 1) for the Type 2 distribution.

Please could someone let me know where my calculations have gone wrong or alternatively help me to interpret why the Type 2 result differs.

Thanks in advance for any help provided.

 

[Julie Lewis]

Suppose that X is Type 2 Geometric, so that P(X=x) = pq^x .

Then:

P(X>=x+k | X>=k) = P(X>=x+k) / P(X>=k)

= {pq^(x+k) + pq^(x+k+1) + ...} / {pq^(k) + pq^(k+1) + ...}

Dividing the numerator and denominator by pq^k gives

{q^x + q^(x+1) + ...} / { 1 + q + ... }

The denominator is equal to 1/(1-q) = 1/p So:

P(X>=x+k | X>=k) = p{q^x + q^(x+1) + ... } = P(X>=x)

 

[Simon C]

Thanks very much Julie. I've followed your proof through and it makes sense.

However one aspect that is still confusing me is that under Type 1 we have proved P(X > x + n | X > n) = P(X > x) whereas under Type 2 we have proved P(X >= x + k | X >= k) = P(X >= x).

I don't think P(X > x + n | X > n) = P(X > x) holds under Type 2. I'm struggling to gain an intuitive understanding of why we are looking at slightly different criteria under the Type 1 and Type 2 scenarios. What is the reason for this?

Thanks again
Simon

 

[Julie Lewis]

Ah, I see your point.

The difference between Type 1 and Type 2 is that, with Type 1 we are counting the number of trials until the first success, and with Type 2 we are counting the number of failures before the first success. So if X is Type 1 Geom, then Y = X-1 is Type 2.

So we have:

P(X > x + n | X > n) = P(X > x)

or equivalently:

P(X-1 > x + n-1 | X-1 > n-1) = P(X-1 > x-1)

Now replacing X-1 by Y, this is:

P(Y > x + n -1 | Y > n-1) = P(Y > x-1)

or:

P(Y >= x+n | Y>=n) = P(Y>= x)

and that's it!

 

[Simon C]

Thanks for that - am glad to see I was on the right track.

Thinking about it further, the different symbols in the formulae make intuitive sense if we consider the following reasoning. For the Type 1 memoryless property, we're considering the probability of there being more than a certain number of trials before the first success. If x equals a particular value under Type 1, then success has occurred at that value so this value of x does not meet the criteria that we are interested in. For a Type 2 variable to meet the memoryless property, we need to make sure that it is considered in a consistent manner to a Type 1 variable. If we are interested in there being more than x trials before the first success under Type 1, this is equivalent to there being more than or equal to x failures under Type 2. Therefore, when considering the memoryless property, it is appropriate and consistent to use > for Type 1 and >= for Type 2.