Expected Mid Change From Poisson Order Flow
In a 1-second interval, buy market orders arrive as $N_b \sim \text{Poisson}(6)$ and sell market orders arrive as $N_s \sim \text{Poisson}(5)$, independently.
Each buy order independently moves the mid price up by $+1$ tick with probability $0.3$ (otherwise no impact). Each sell order independently moves the mid price down by $-1$ tick with probability $0.25$ (otherwise no impact). Impacts are independent across all orders and sides.
Let $M$ be the total mid price change (the sum of all individual impacts).
- Compute $E[M]$ and $\text{Var}(M)$.
- Using a normal approximation, estimate $P(M \ge 2)$. State the mean and variance you plug in.
Hints
- When a Poisson number of i.i.d. Bernoulli trials occur, the count of successes has a nice closed-form distribution -- what is it?
- Apply Poisson thinning: $\text{Poisson}(\lambda)$ trials each succeeding with probability $p$ gives $\text{Poisson}(\lambda p)$ successes. The total mid change is the difference of two independent thinned Poisson variables.
- Compute $E[M] = 6(0.3) - 5(0.25)$ and $\text{Var}(M) = 6(0.3) + 5(0.25)$, then standardize $M$ as a normal with these parameters to approximate the tail probability.
Worked Solution
How to Think About It: This is a compound Poisson problem dressed up in order-flow language. Each side generates a random number of orders, and each order independently either moves the mid or does nothing. The key shortcut is Poisson thinning: if you take a $\text{Poisson}(\lambda)$ number of i.i.d. coin flips, the number of successes is itself Poisson. That collapses the whole problem into the difference of two independent Poisson random variables, which is quick to work with.
Quick Estimate: Buy-side impact count is $\text{Poisson}(6 \times 0.3) = \text{Poisson}(1.8)$, sell-side impact count is $\text{Poisson}(5 \times 0.25) = \text{Poisson}(1.25)$. So the expected net move is $1.8 - 1.25 = 0.55$ ticks upward, and the variance is $1.8 + 1.25 = 3.05$. Standard deviation is about $1.75$ ticks. The mid drifts slightly up on average but with plenty of noise -- a $\pm 2\sigma$ range of roughly $-2.9$ to $+4.0$ ticks.
Approach: Use Poisson thinning, then properties of sums of independent random variables.
Formal Solution:
Define the buy-side impact as $B = \sum_{i=1}^{N_b} X_i$ where each $X_i \sim \text{Bernoulli}(0.3)$ takes value $+1$ with probability $0.3$ and $0$ otherwise. Define the sell-side impact as $S = \sum_{j=1}^{N_s} Y_j$ where each $Y_j \sim \text{Bernoulli}(0.25)$ takes value $1$ with probability $0.25$ and $0$ otherwise. Then $M = B - S$.
By Poisson thinning:
$$B \sim \text{Poisson}(\lambda_b p_b) = \text{Poisson}(6 \times 0.3) = \text{Poisson}(1.8)$$
$$S \sim \text{Poisson}(\lambda_s p_s) = \text{Poisson}(5 \times 0.25) = \text{Poisson}(1.25)$$
Since $N_b$ and $N_s$ are independent, $B$ and $S$ are independent.
Part 1: Mean and Variance
$$E[M] = E[B] - E[S] = 1.8 - 1.25 = 0.55$$
$$\text{Var}(M) = \text{Var}(B) + \text{Var}(S) = 1.8 + 1.25 = 3.05$$
The variance adds because $B$ and $S$ are independent (the minus sign does not matter for variance).
Part 2: Normal Approximation for $P(M \ge 2)$
Approximate $M \sim N(0.55, 3.05)$, so $\sigma = \sqrt{3.05} \approx 1.747$.
Without continuity correction:
$$P(M \ge 2) \approx P\!\left(Z \ge \frac{2 - 0.55}{1.747}\right) = P(Z \ge 0.830) \approx 1 - \Phi(0.830) \approx 0.203$$
With continuity correction (since $M$ is integer-valued):
$$P(M \ge 2) = P(M \ge 1.5) \approx P\!\left(Z \ge \frac{1.5 - 0.55}{1.747}\right) = P(Z \ge 0.544) \approx 1 - \Phi(0.544) \approx 0.293$$
The continuity-corrected version is more accurate here.
Answer: $E[M] = 0.55$ ticks, $\text{Var}(M) = 3.05$, and $P(M \ge 2) \approx 0.29$ (with continuity correction) or $\approx 0.20$ (without).
Intuition
This problem illustrates Poisson thinning, one of the most useful tricks in stochastic modeling for markets. When arrivals are Poisson and each arrival independently triggers an event with some probability, the triggered events are also Poisson with a scaled rate. This lets you collapse a two-layer randomness (random number of orders, then random impact per order) into a single-layer problem. In real market microstructure, this decomposition shows up constantly: order arrivals are roughly Poisson, and each order's impact depends on size, aggression, and book state. The net flow over short intervals is well-modeled as the difference of two independent Poisson streams, giving you fast formulas for the mean drift and volatility of the mid.
The practical takeaway: the expected mid change per second is just $\lambda_b p_b - \lambda_s p_s$, and the variance is $\lambda_b p_b + \lambda_s p_s$. Notice the variance always adds regardless of direction -- buy pressure and sell pressure both contribute to uncertainty. This is why even in a "balanced" market where the expected drift is near zero, the mid still jiggles around. In an interview, leading with the thinning argument and stating the mean and variance in one line shows you understand the structure rather than grinding through conditional expectations.