Question

A drunkard executes a “random walk” in the following way: each minute he takes a step north or southm with probability 1/2 each, and his successive step directions are independent. His step lenght is 50 cm. Us the central limit theorem to approximate the probability distribution of his location after 1h. Where is he most likely to be?

Answer 1

Assume that a step to the north is in the positive direction and a step in the south is in the negative direction.

Then,

$E(X_i)=50\left(\frac{1}{2}\right)-50\left(\frac{1}{2}\right)\\ \\=25-25=0$

and

$\sigma^2(X_i)=E(X_i^2)-E(X_i)^2 \\ \\ =50^2\left(\frac{1}{2}\right)+50^2\left(\frac{1}{2}\right)-0^2=2500\left(\frac{1}{2}\right)+2500\left(\frac{1}{2}\right)-0 \\ \\ =2500 \\ \\ \Rightarrow\sigma(X_i)=50$

After 1 hour of independent steps,

$E(X)=\sum_{i=1}^{60}E(X_i)=60(0)=0$

and

$\sigma^2(X)=\sum_{i=1}^{60}\sigma^2(X_i) \\ \\ =60(50)=300 \\ \\ \Rightarrow\sigma(X)=\sqrt{300}=17.32$

Therefore, by Central Limit Theorem, the distribution of the random walk is normal with a mean of 0 cm and a standard deviation of 17.32 cm. He is expected to be somewhere around his starting point after 1 hour.

Answer 2

You should get the mean through:

Mean= 50 (1/2)- 50(1/2) which will give you zero.

Second, you will get the standard deviation.

Then multiply it 50 by 60

So:

50 (60) = 300

Square root of 300 = 17.32.

Through this, you can say that he is just near to the place where he is walking after 1 hour.