Binomial Expected value and Variance

Question

I'm seeking basic analytical convergence between Black-Scholes and Binomial model. I've been following the Feng et al. (2012) approach. However, we know that $E(k)= Nq$ and $Var(k)= Nq(1-q)$. I've previously calculated that \begin{equation} \ln\left(\frac{S^{N}_{k}}{S^{0}_{0}}\right)=\ln\left(u^{k}v^{N-k}\right)=k\ln u+(N-k)\ln v=k\ln\left(\frac{u}{v}\right)+N\ln v. \end{equation} Here, $\ln\left(\frac{S^{N}_{k}}{S^{0}_{0}}\right)$ is equivalent to a continuously compounded return over the $N$ periods. If we proceed to find the expected value and variance, we obtain $$ E\left[\ln\left(\frac{S^{N}_{k}}{S^{0}_{0}}\right)\right]=E\left[k\ln\left(\frac{u}{v}\right)+N\ln v\right] =Nq\ln\left(\frac{u}{v}\right)+N\ln v=\left(q\ln u+(1-q)\ln v\right)N $$ and $$ \text{Var}\left[\ln\left(\frac{S^{N}_{k}}{S^{0}_{0}}\right)\right]=\text{Var}\left[k\ln\left(\frac{u}{v}\right)+N\ln v\right] \\$$ I am now stuck with how to proceed. If I substitute in our $Var(k)=Nq(1-q)$ expression i end up with (i believe) $$ \text{Var}\left[\ln\left(\frac{S^{N}_{k}}{S^{0}_{0}}\right)\right]=Nq(1-q)\ln\left(\frac{u}{v}\right)+N\ln v. $$ However, the article gives $\text{Var}\left[\ln\left(\frac{S^{N}_{k}}{S^{0}_{0}}\right)\right]=q(1-q)\left(\ln\left(\frac{u}{v}\right)\right)^{2}N.$ Im unsure how to obtain this result from what i have, i'd really like to know how to obtain it, however my distribution theory/knowledge is rather basic! Any help would be greatly appreciated, kind regards.

Answer 1

When you multiply a random variable $X$ by a constant $a$, its variance gets multiplied by $a^2$, whereas adding a constant $b$ to it doesn't change the variance: $$\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X).$$ For your problem, $X=k$ is the random variable and $a=\ln\left(\frac uv\right)$ and $b=N\ln v$ are the constants. So $$\text{Var}\left[k\ln\left(\frac{u}{v}\right)+N\ln v\right] =\left(\ln\left(\frac uv\right)\right)^2 \mathrm{Var}(k) =\left(\ln\left(\frac uv\right)\right)^2 Nq(1-q). $$