Consider the Poisson branching model with mean $c = 1$ and root Eve. For $n ≥ 3$, let $A_n$ be the event where Eve has precisely two children, Dana and Fan, and that the total tree size $T = n$. Let X be the size of the subtree with root Dana. For each $j \geq 1$, find $\lim_{n \to \infty} \mathbb{P}(X=j|A_n)$. Find the asymptotic formula for $\mathbb{P}(\frac{n}{3}<X<\frac{2n}{3})$.
$\mathbf{Hint:}$ Set $x=n\alpha$ and $y=n-1-x \sim n\beta$ with $\alpha+\beta = 1$. Use the asymptotic equation $\mathbb{P}(T=k) = \frac{e^{-k}k^{k-1}}{k!} \sim \frac{1}{\sqrt{2\pi}}k^{-\frac{3}{2}}$ to estimate $\mathbb{P}(X=x)\mathbb{P}(Y=y)$. The denominator $\mathbb{P}(X+Y = n-1)$ is dominated by $min(X,Y)$ being small. Perhaps surprisingly, the conditional distribution of $X$ is highly skewed to the corners.
$\mathbf{Idea:}$ That's what I managed to do.
$$\mathbb{P}(X=j|A_n)= \frac{\mathbb{P}(X=j \cap A_n)}{\mathbb{P}(A_n)}=\frac{\mathbb{P}(X=j \cap Z_1=2 \cap T=n)}{\mathbb{P}(Z_1=2 \cap T=n)}$$
Then
$$\mathbb{P}(A_n)= \mathbb{P}(Z_1=2 \cap T=n)= \mathbb{P}(T=n|Z_1=2)\mathbb{P}(Z_1=2)$$
$$\mathbb{P}(X=j \cap Z_1=2 \cap T=n)= \mathbb{P}(X=j \cap T=n|Z_1=2)\mathbb{P}(Z_1=2)$$
We denote by $Y$ the size of the subtree with root Fan, then
$$\mathbb{P}(X=j \cap T=n|Z_1=2) = \mathbb{P}(X=j \cap Y=n-1-j|Z_1=2)= \\
\mathbb{P}(X=j \cap Y=n-1-j)=\mathbb{P}(X=j)\mathbb{P}(Y=n-1-j).$$
As the hint suggest we use the asymptotic equation to estimate $\mathbb{P}(X=j)\mathbb{P}(Y=n-1-j)$. Then
$$\mathbb{P}(X=j) \sim \frac{1}{\sqrt{2\pi}}j^{-\frac{3}{2}} = \frac{1}{\sqrt{2\pi}}(n\alpha)^{-\frac{3}{2}}$$
and
$$\mathbb{P}(Y=n-1-j) \sim \frac{1}{\sqrt{2\pi}}(n-1-j)^{-\frac{3}{2}} = \frac{1}{\sqrt{2\pi}}(n\beta)^{-\frac{3}{2}}$$
Combining them both we get:
$$\mathbb{P}(X=j)\mathbb{P}(Y=n-1-j) \sim \frac{1}{\sqrt{2\pi}}(n\alpha)^{-\frac{3}{2}}(n\beta)^{-\frac{3}{2}}=\frac{1}{\sqrt{2\pi}}n^{-3}(\alpha\beta)^{-\frac{3}{2}}$$
Finally we have that
$$\mathbb{P}(A_n) =\mathbb{P}(T=n|Z_1=2)\mathbb{P}(Z_1=2)= \mathbb{P}(X+Y=n-1)\mathbb{P}(Z_1=2)$$
And we can approximate $\mathbb{P}(X+Y=n-1) \leq min(X,Y)$.
Then we can conclude that
$$\mathbb{P}(X=j|A_n)=\frac{\mathbb{P}(X=j \cap T=n|Z_1=2)}{\mathbb{P}(T=n|Z_1=2)} \geq \frac{\mathbb{P}(X=j)\mathbb{P}(Y=n-1-j)}{min(X,Y)} \sim \frac{\frac{1}{\sqrt{2\pi}}n^{-3}(\alpha\beta)^{-\frac{3}{2}}}{min(X,Y)}$$
$\mathbf{Questions}$ I'm confused about this $min(X,Y)$. I don't understand if I can conclude like this the first part of the exercise. And I actually have no idea how to answer the second part (Find the asymptotic formula for $\mathbb{P}(\frac{n}{3}<X<\frac{2n}{3})$).
Thank you all for the help!