A Few Random Walk Questions

[HP]

Okay, so you've got your standard random walk:

Flip a coin. If heads, move one unit right, if not, one unit left. Start at x=0

The chance you will eventually get to x=1 approaches 1 as the number of additional flips you are willing to take approaches infinity, correct?

And again the chance you will eventually make it to say x=100 (or any other whole number) approaches 1 as the number of flips you are willing to take approaches infinity, correct?

I'm pretty sure the answer is yes and yes so far. How about this though:

You start off at x=1

Flip a coin. If heads, increase value by 10%. If tails, decrease value by 10%. Given you are prepared to flip forever, what are the chances you will make it to x=1.1 or higher? Is it 100% ?

If it is 100%, what if you start at x=1 and use a log-normal distribution (which yields an average change of 0) to figure out where your next step lands you. Will you always make it to a value greater than x=1 eventually ?

[sixhigh]

If I understand you correctly you mean the following: If the coin comes heads the new value (x_(n+1)) will be x_n*1.10, if the coin comes tails the new value will be x_n*0.90.

In this case the probability of reaching any point above the initial value will be less than 100%. That is because the whole process is a supermartingale - i.e. the expected value of x_(n+1) is lower than the ev of x_n

E[x_(n+1) | x_n] = 1/2 * x_n * 1.1 + 1/2 * x_y * 0.9 = 0.99 * x_n < x_n

thus E[x_n] --> 0 (for n --> \infty)

Now there are ways to calculate the exact probability of x_n reaching a certain value at least once, but iirc this is quite difficult.

I don't know what you mean by using a certain distribution to determine x_(n+1). Should the probability of moving up/down be based on the value of x?

And the answer to the first questions is yes - you will reach every value infinitely often in a one-dimensional random walk.

[HP]

thanks for the reply

[ QUOTE ]
1/2 * x_n * 1.1 + 1/2 * x_y * 0.9 = 0.99 * x_n

[/ QUOTE ]
Am I missing something here? why is it 0.99 instead of 1?

[sixhigh]

Whops - that's probably wrong. The entire post.

E[x_n] = 1 for all n,

but still x_n --> 0. (p-almost sure)

Say you walk n steps and k times the coin lands on heads. The value of x_n then will be 1 * 1.1^k * 0.9^(n-k).
For this value to be greater than (say) x_0 = 1 we get the following expression:

1.1^k*0.9^(n-k) > 1

Solving this expression for k (and using Derive) we get

k > n * ln(1/0.9) / ln(1.1/0.9)

or

k > n * 0.525

Now since k has binomial distribution (with the parameter p=1/2) we know by the law of large numbers that it will (p-almost surely) not be in this region for large n.

The EV of x_n being 1 although x_n almost surely approaches 0 follows for nearly the same reasons the St.-Petersburg paradox does.

Now since the random variable approaches zero almost surely it will almost surely never hit a certain value above 0 infinitely often. Thus (i guess) it will hit no value above 0 with probability 1.

[HP]

cool, thanks, makes sense to me

[HP]

I'm still uneasy about this all however

I had the money markets in mind when I made the OP

Let's assume an efficient market (random walk)

And let's simplify stuff and say the interest rates in the USA and Europe will always be the same. If you don't know why this matters don't worry

so, With similar logic to the above you could say that looking at the USD vs the EUR, through a random walk the USA will approach 0

But then the same could be said for the EUR vs the USD

I'm still pondering about what kind of probability density functions we have for the change of USD/EUR after a day

hmm...