0
$\begingroup$

A particle starts at the origin of the real line and moves along the line in jumps of one unit. For each jump the probability is $p$ that the particle will jump one unit to the left and the probability is $1 − p$ that the particle will jump one unit to the right. Let $X_n$ be the position of the particle after $n$ jumps. Find $\mathbb{E}[X_n]$ and $\mathbb{V}[X_n]$. (This is known as a random walk.)

For this problem, I am not sure if I should use the binomial random variable to find the expectation or find the expectation directly by summing the expected values for all $n$ steps.

Since it's a sequence of repeated trials of probability $p$ of moving one step to the left and $1-p$ of moving one step to the right. It seems from intuition that the expected value would be

$$\sum_{i=0}^n (-1)(p)+(1)(1-p)\\=(1-2p)n$$ and the variance calculated from $\mathbb{E}[X^2]-\mathbb{E}[X]^2$. So, $$\sum_{i=0}^n (-1)^2(p)+(1)^2(1-p)\\=n\\\mathbb{V}[X_n]=n-(n-2np)^2$$

But I am not sure if this approach is correct?

$\endgroup$
1
  • 1
    $\begingroup$ hint: $X_n=Y_1+Y_2+\ldots +Y_n$ where the $Y_k$ are iid rv with Distribution $P(Y_1=-1)=p$ and $P(Y_1=1)=1-p$ $\endgroup$
    – Masacroso
    Commented Sep 22, 2022 at 14:57

1 Answer 1

1
$\begingroup$

No need to use intuition. Let $Y_n$ be the count of left steps upto time $t$. Then $Y_n$ follows a Binomial $(n, p)$ distribution.

And $X_n = n-2 \, Y_n $

Can you go on from here?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .