I already asked this yesterday at "Economics Stack Exchange" but think this question might be better suited here. In the meantime i really tried to solve it by myself, but couldn't find anything what might help me. It's not just about the solutions, i do really want to understand how to solve problems like this.

"I am currently studying for my upcomming exams. There is an exercise i can't solve or even understand properly.

The full exercise is: "You bought 100 shares of company A and 200 shares of company B. The shares of A are bid \$50 and ask \$60, while the shares of B are bid \$25 and ask \$35. The bid-ask spreads of both A and B are normally distributed with mean \$10 and standard deviation \$3.

Determine the distributions of the proportional bid-ask spreads for A and B."

I already got the proportional bid-ask-spread for A and B by the formula $s_{p}(X) = \frac{ASK - BID}{MEAN}$. Therefore $s_{p}(A) \approx 0.18$ and $s_{p}(B) \approx 0.33$.

Now i need to calculate the distributions of those spreads. (The actual aim of this exercise is to calculate the cost of liquidation in a stressed market.)

I'm not quite sure what is meant by "distribution", so i assume it's the mean and standard deviation of those spreads. I just can't get my head around the standard deviation, since i need at least two values to calculate the standard deviation. (as far as i undersand) But i don't have more than one value for each spread.

How do i have to solve this exercise? Like, is there a general way of doing it?"

UPDATE: I need these results to calculate the "cost of liquidation in a stressed market". I read that one needs to use the following formula for this.

$\sum_{i=1}^{n} \frac{1}{2}(\mu_i + \lambda_i \sigma_i)\alpha_i$, where

$n=2$,

$\mu_i = s_p(X_i)$, so in my case $\mu_1 = 0.18$ and $\mu_2 = 0.33$,

$\alpha_i = \text{volume } X_i$, in my case $\alpha_1 = 100$ and $\alpha_2 = 200$,

$\lambda_i = \text{confidence-level}$, like $\lambda_1 = \lambda_2 = 2.33$ for $99\%$ confidence-level and finally

$\sigma_i = \text{that "distribution" (standard-deviation?) value i can't calculate}$.

Maybe this can describe my problem in more detail.

• Hi nobody, welcome to Quant.SE! I've left a note on your question on Economics as to prevent a duplication of efforts. Is your formula for $s_p(X)$ wrong, shouldn't it be as described here: investopedia.com/terms/p/proportional-spread.asp ? Feb 23 '16 at 20:54
• Hi Bob Jansen, i just read your comment. I tought i would leave the question there with an link to this post in case anyone find's it and needs an answer like me. Or should i delete the question on Economics? EDIT: just deleted the question on Economics. Feb 23 '16 at 20:56
• Hi Bob Jansen, i used that formula, since i calculated the MEAN by (ASK+BID)/2. What i just don't understand is what is meant by "distribution" and how to calculate said values. Feb 23 '16 at 20:59
• Ah OK, I thought $\mathrm{MEAN}$ was the mean of the normal distribution in the previous paragraph. Feb 23 '16 at 21:01

You can not derive the distribution of proportional spread with the information given in your question. You have given $S_p{(A)}$ and $S_p (B)$. By assuming equal probabilities for both, you can simply calculate standard deviation of proportional spread as: $$Var(S_p)=E[(S_p-\mu)^2]$$ So, $\mu = 0.5*0.18 + .5*.33 = 0.255$, and $$Var(S_p)=.5(0.18-.255)^2 + 0.5(.33-.255)^2=0.005625$$ $$\sigma=\sqrt{Var(S_p)}=0.075$$
• Thanks a lot!! Just to not miss anythng.. so in my case $\sigma_1 = \sigma_2 = 0.075$ for that formula in my initial post (liquidation in stressed market)? Feb 24 '16 at 18:23
• @nobody one thing is sure, you can not derive the distribution of proportional spread with the information given in your question. The spread of both the company A and B has identical distribution. So, if you assume that proportional spread too has identical distribution, then you can use same standard deviation for both. As an alternative, you can use $|.18-.255|$ as proportional spread standard deviation for A and $|.33-.255|$ as proportional spread standard deviation for B. Feb 24 '16 at 18:46