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The means add regardless of any other assumptions. Assuming (as usual in random walks) that the increments are independent, the variances also add. Hence the variance at time $t$ is $\sigma^2t$, and the mean is $\mu t$. Now $$\text{Var}(x_t)=E[x_t^2]-(E[x_t])^2=E[x_t^2]-\mu^2t^2=\sigma^2t$$ hence $$E[x_t^2]=\sigma^2t + \mu^2t^2.$$


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Then after the mosquito goes to $0$, the mosquito must immediately go back to $1$ next move, i.e. the whole situation restarts. Consider the case that when the mosquito goes from $1$ to $4$ without passing $0$ as a success, and the case when the mosquito goes from $1$ to $0$ without passing $4$ as a failure. You already calculated the two probabilities as ...


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Let $P(i)$ denote the probability he'll get to $\$1,000,002$ if he starts from $\$i$. We know that $$P(0)=0\;\;\;P(1,000,002)=1$$ You are asked to find $P(1,000,000)$. Imagine he has $\$i$ and places a bet. He either gets to state $(i+1)$ or to state $(i-1)$ and we have: $$P(i) = \frac 13 P(i+1)+\frac 23 P(i-1)$$ It is easier to start near the "ruin ...


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You haven't formalized what you mean by "the result is symmetric". The individual graphs are certainly not symmetric, you haven't specified a distribution whose symmetry we could inquire into, and if you had, it wouldn't be clear whether you could judge its symmetry by examining a couple of samples with the naked eye. So presumably what you mean is roughly ...


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Well, I managed to hand-build a dirty distribution that works for me. The idea is to paste two geometric distributions together, and to truncate them not to get out of $S$: give yourself two parameters $k \in [0,\frac{2}{3}]$ (overload for the probability of "staying here") and $\lambda \in \mathbb{R}^+$ (parameter for the geometric distribution). compute ...


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Your chain is recurrent (since it is a finite state makov chain). In this context there is a general formula that relates the invariant measure $\nu(\cdot)$ and the expected time of return $\Bbb{E}[\tau_\cdot]$. Let $X_0 = i$, define $T_1 = \inf\{k>0, X_k = i\}, T_2 = \inf\{k> T_1, X_k = i\}, \ldots$ Those $T_j$ are the times of first return to the ...



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