Variance of a compound daily PnL

Question

A desk's number of fills in a day is N \sim 	ext{Poisson}(20). Each fill's PnL is independent with mean \mu = \50 and standard deviation \sigma = \500 (so variance 250{,}000), independent of N. The day's total PnL is Y = \sum{i=1}^{N} Xi. Use the law of total variance to find \operatorname{Var}(Y).

Accepted Answer

Condition on N. Given N = n, the total is a sum of n i.i.d. fills, so \mathbb{E}[Y \mid N] = N\mu, \qquad \operatorname{Var}(Y \mid N) = N\sigma^2. The law of total variance splits the answer into within and between parts: \operatorname{Var}(Y) = \underbrace{\mathbb{E}[\operatorname{Var}(Y \mid N)]}{	ext{within}} + \underbrace{\operatorname{Var}(\mathbb{E}[Y \mid N])}{	ext{between}} = \mathbb{E}[N]\sigma^2 + \operatorname{Var}(N)\mu^2. For a Poisson, \mathbb{E}[N] = \operatorname{Var}(N) = 20. Plugging in: \operatorname{Var}(Y) = 20 \cdot 250{,}000 + 20 \cdot 50^2 = 5{,}000{,}000 + 50{,}000…

Variance of a compound daily PnL

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