Let $N$ be a Poisson random variable with parameter $\lambda$. If the parameter $\lambda$ is not fixed, but an exponential random variable with parameter $1$, find $E[N]$.
Here is a correct solution given to this problem :
$E[N] = E[E[N | \Lambda]] = \int_{0}^{\infty} E[N|\Lambda = \lambda] \cdot f_{\Lambda}(\lambda)d\lambda = \int_{0}^{\infty} \lambda \cdot e^{-\lambda} = \Gamma(2) = 1$, where $\Lambda$ represents an exponential random variable with parameter $1$.
Here is my proposed solution :
$E[N] = E[E[N | \Lambda = \lambda]] = E[\lambda] = 1$.
Basically, I don't see why the integration in the correct solution is necessary - we already have the necessary knowledge about the desired expectations here, since a Poisson random variable with parameter $\lambda$ has expectation $\lambda$, and an exponential random variable with parameter $\theta$ has expectation $\frac{1}{\theta}$.
Have I violated the theory of conditional expectation here ?
Thanks!
As long as you're familiar with expectations or any moment you are interested in of typical random variables, you don't have to do the computation all the time.
Note that we have $N \sim Poiss(\Lambda)$, where $\Lambda \sim \mathcal Exp(1)$
Hence we have $\mathbb E[N] = \mathbb E[ \mathbb E[N|\Lambda] ] = \mathbb E[\Lambda] = \frac{1}{1} = 1$, since $\mathbb E[Poiss(x)] = x$ and $\mathbb E[\mathcal Exp(x)] = \frac{1}{x}$, when $x \in \mathbb R_+$.
As for your second question, it isn't so good to write it $\mathbb E[N] = \mathbb E[ \mathbb E[N | \Lambda = \lambda]]$
You may as why? Conditional Expectation is a random variable, so $\mathbb E[X|Y]:\Omega \to E$, where $\Omega, E$ are sets. It is often written in a way $\mathbb E[X|Y](\omega) = \mathbb E[X|Y=Y(\omega)]$. Sometimes conditional expectation is considered as function $E \to E$ in a form $\mathbb E[X|Y](y) = \mathbb E[X|Y=y]$.
In both of those cases, when we consider object of form $\mathbb E[X|Y="something"]$ it isn't random anymore, so taking expectation of it will give you the same thing. You say they said $\lambda$ isn't fixed, said that it is an exponential random variable. But you considered that $N \sim Poiss(\Lambda)$ So what is $\lambda$ then and what condition $\Lambda = \lambda$ mean? That can definitelly mean something, but we would need some information about the joint distribution of $(\Lambda,\lambda)$.
Long story short: You can use and write $\mathbb E[N|\Lambda =\lambda]$ when $\lambda$ is fixed as to COMPUTE conditional expectation $\mathbb E[N|\Lambda]$, but when you want to use total expectation property, you need to have $\mathbb E[N|\Lambda]$-random variable and not $\mathbb E[N | \Lambda = \lambda]$ - fixed number