Consider part (ii) of the below question: question

Note: $p$ denotes the probability of an upward move of $1$. $q$ denotes the probability of a downward move of $1$.

Having looked as the solutions (which do not include workings), I know that the solution is $0.97831$.

Clearly, this is calculated using the standard normal distribution. I have attempted this as follows...

The expected value of the process after 1 move is $$ \mathbb{E}[Z_n] = (1 \cdot p) + (-1 \cdot q) = 0.45 - 0.55 = -0.1 $$ and the variance of the $Z_n$ variables is given by $$ Var(Z_n) = 4pq = 0.99 = 0.995^2 $$ So, we have deduced that $Z_n ~ N(-0.1, 0.995^2)$ and thus $Y_{80} ~ N(-8, 0.995^2)$. Then, using the standard normal distribution, we have $$ P(Y_{80} \leq 10) = P(z \leq \frac{x-\mu}{\sigma}) \\ = P \left( z \leq \frac{10-(-8)}{0.995} \right) \\ = P(z \leq 18.091) $$ This is clearly the point at which I have gone wrong, but I cannot see why. Can anyone please explain this to me?


1 Answer 1


Hint: if $X\sim \mathcal{N}(m_x, \sigma_x^2)$ and $Y\sim \mathcal{N}(m_y, \sigma_y^2)$ and they are independent, then $$X+Y \sim \mathcal{N}(m_x+m_y, \sigma_x^2+\sigma_y^2).$$

  • 1
    $\begingroup$ Ah, I see! My value for the standard deviation was incorrect. Thanks. $\endgroup$
    – M Smith
    Commented May 21, 2017 at 19:00

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