Asymptotic distribution of sample mean of the sum of two poisson distribution

I'm trying to calculate the asymptotic distribution of the sample mean of the sum of two Poisson distributions.

Sample 1 is of size N1, and is from a Poisson distribution with expectation $$\mu_1$$. Sample 2 is of Size N2, and is from a Poisson distribution with expectation $$\mu_2$$.

Now, I was able to derive that when $$N_1 \uparrow \infty$$, and $$N_2 \uparrow \infty$$ at the same rate, the sample mean of the sum of two samples will be $$\frac{(N_1 \cdot \mu_1 + N_2 \cdot \mu_2)}{N_1 + N_2}$$.

From here, I'm having trouble deriving the standard deviation of the asymptotic distribution of the sample mean. How would I go about it?

• Are they independent? You can't say anything about the standard deviation of the sum if they're not independent. If they are, then use the fact that $\text{Var}(X+Y)=\text{Var}(X)+\text{Var}(Y)$ when $X$ and $Y$ are independent. (Note variance is the square of standard deviation.) – kccu Apr 17 at 3:09
• " a Poisson distribution with expectation $\mu_1$ and standard deviation $\sigma_1$" Sounds weird to specify two parameters, a Poisson distribution has a single parameter and the expectation equals the variance. You got that right? – leonbloy Apr 17 at 12:19
• @kccu Yes, they are independent. In that case, does Var(X+Y) = Var(X) + Var(Y) also apply to asymptotic distribution of the sample mean? I think what's unclear to , me is the difference between the Poisson distribution and the distribution of the sample mean. – kunichi_kimura Apr 17 at 13:16
• @leonbloy Yes, I made edits accordingly. But the question is about the distribution of the sample mean. Shouldn't the sample mean asymptotically normally distributed? So it won't have variance that equals to the expectation – kunichi_kimura Apr 17 at 13:17

Let $$X_1,X_2,\dots,X_{N_1}$$ be the samples from the first Poisson distribution, which are independent, and let $$S_{N_1}$$ be their sum. Let $$Y_1,Y_2,\dots,Y_{N_2}$$ be the samples from the second Poisson distribution, which are also independent (and independent of the $$X_i$$'s), and let $$T_{N_2}$$ be their sum. The $$X_i$$'s all have mean and variance $$\mu_1$$ and the $$Y_i$$'s all have mean and variance $$\mu_2$$.
The sample mean is then the random variable: $$\frac{X_1+\cdots+X_{N_1}+Y_1+\cdots+Y_{N_2}}{N_1+N_2} = \frac{S_{N_1}+T_{N_2}}{N_1+N_2}.$$ You have found the mean of the sample mean (not the sample mean, which is again a random variable) to be $$E\left[\frac{X_1+\cdots+X_{N_1}+Y_1+\cdots+Y_{N_2}}{N_1+N_2}\right]=\frac{E[S_{N_1}]+E[S_{N_2}]}{N_1+N_2}=\frac{N_1 \mu_1+N_2\mu_2}{N_1+N_2}.$$ To find the standard deviation of the sample mean, first find the variance. The two facts you need to use are:
1. $$\text{Var}(aX)=a^2\text{Var}(X)$$ for any constant $$a$$ and random variable $$X$$.
2. If $$X$$ and $$Y$$ are independent, then $$\text{Var}(X+Y)=\text{Var}(X)+\text{Var}(Y)$$.
Thus, \begin{align*} \text{Var}\left(\frac{X_1+\cdots+X_{N_1}+Y_1+\cdots+Y_{N_2}}{N_1+N_2}\right) &=\left(\frac{1}{N_1+N_2}\right)^2\text{Var}(X_1+\cdots+X_{N_1}+Y_1+\cdots+Y_{N_2})\\ &=\cdots (\text{use fact 2 now}) \end{align*} I'll let you finish from here.