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?
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?