泊松分布矩估计法：二手车销售场景下μ的矩估计求解问询

Hey there, let's break down how to calculate the method of moments (MoM) estimator and maximum likelihood estimator (MLE) for μ in this truncated Poisson scenario. The key twist here is that we don't have full daily sales data—only counts of days with 0 sales vs. at least 1 sale, so we need to adjust our usual estimation approaches accordingly.

We can't use the standard first sample moment (sample mean of daily sales) because we don't know the exact sales values for the 10 days with at least 1 sale—we only know they're ≥1. Instead, we need to use a moment that we can actually estimate from the observed data.

Let's define an indicator random variable ( Y = 1 ) if daily sales ( X \geq 1 ), and ( Y = 0 ) if ( X = 0 ). The population moment for ( Y ) is:
[ E[Y] = P(X \geq 1) = 1 - P(X=0) = 1 - e^{-\mu} ]

The sample moment for ( Y ) is the proportion of days with at least 1 sale, which is ( 10/30 = 1/3 ). MoM tells us to set the population moment equal to the sample moment:
[ 1 - e^{-\hat{\mu}_{\text{MoM}}} = \frac{1}{3} ]

Solving for ( \hat{\mu}{\text{MoM}} ):
[ e^{-\hat{\mu}{\text{MoM}}} = \frac{2}{3} ]
[ \hat{\mu}_{\text{MoM}} = -\ln\left(\frac{2}{3}\right) \approx 0.4055 ]

Alternatively, we could match the sample proportion of days with 0 sales (( 20/30 = 2/3 )) to the population probability ( P(X=0) = e^{-\mu} ), which gives the exact same equation and result.

To address your initial thought: you mentioned setting the first sample moment ( M_1 \geq 1/3 )—that's the lower bound of the sample mean (since each of the 10 days has at least 1 sale), but that's not a valid MoM estimator. MoM requires equating a known sample moment to its corresponding population moment, and since we don't have the full sample mean, we have to use the observable proportion of 0/non-0 sales instead.

Now let's tackle the MLE. The likelihood function is built directly from the observed data: 20 days with 0 sales, 10 days with at least 1 sale.

For a single day with 0 sales, the probability is ( P(X=0) = e^{-\mu} ). For a single day with at least 1 sale, the probability is ( P(X \geq 1) = 1 - e^{-\mu} ). So the likelihood function for 30 days is:
[ L(\mu) = \left(e^{{-\mu}\right)}{20} \times \left(1 - e^{{-\mu}\right)}{10} ]

Take the natural logarithm to simplify differentiation (this gives the log-likelihood, which is easier to work with):
[ \ln L(\mu) = -20\mu + 10\ln\left(1 - e^{-\mu}\right) ]

Now take the derivative with respect to ( \mu ) and set it to 0 to find the value that maximizes the likelihood:
[ \frac{d}{d\mu} \ln L(\mu) = -20 + 10 \times \frac{e^{-\mu}}{1 - e^{-\mu}} = 0 ]

Solving for ( \mu ):
[ 20 = 10 \times \frac{e^{-\mu}}{1 - e^{-\mu}} ]
[ 2 = \frac{e^{-\mu}}{1 - e^{-\mu}} ]
[ 2(1 - e^{-\mu}) = e^{-\mu} ]
[ 2 - 2e^{-\mu} = e^{-\mu} ]
[ 2 = 3e^{-\mu} ]
[ e^{-\mu} = \frac{2}{3} ]
[ \hat{\mu}_{\text{MLE}} = -\ln\left(\frac{2}{3}\right) \approx 0.4055 ]

A fun quirk here: in this specific truncated scenario, the MoM and MLE end up being identical because we're basing both on the same observable proportion data.

内容的提问来源于stack exchange，提问作者Eoin