Let $Y_1,Y_2,Y_3,...,Y_n$ be a random sample from the exponential distribution having PDF $f(y;\lambda)= \lambda e^{-y\lambda},$ $y>0.$
A) Show that $\hat\lambda_n = Y_1$ is not consistent for $\lambda$.
B)Show that $\hat\lambda_n =\sum\limits_{i=1}^n Y^2_i$ is not consistent for $\lambda$.
The hint in the back of the book has probably confused me but it says to use $P(Y_1 > 2\lambda)$ and I don't see where that is coming from. I was assuming that I was going to use an epsilon-delta proof, or use chebyshev's but now I'm just confused.
A consistent estimator will converge in probability to the parameter it estimates, for increasing sample size. Clearly our intuition says that $\hat \lambda_n = Y_1$ will not: the first observation of a sample of size $5$ is no different than the first observation of a sample of size $5000000$. So, you can see that $\Pr[Y_1 > 2\lambda]$ is a constant function of $n$, which contradicts consistency: if $\hat \lambda_n$ were consistent, $$\lim_{n \to \infty} \Pr[|\hat\lambda_n - \lambda| \ge \epsilon] = 0$$ for any $\epsilon > 0$.
With the above in mind, I leave part (B) to you to consider further.
Okay, so if you want an explicit proof: Let $\hat \lambda_n = Y_1$. Then for any $\epsilon \in (0,\lambda]$, $$\begin{align*} \lim_{n \to \infty} \Pr[|\hat \lambda_n - \lambda| \ge \epsilon] &= \lim_{n \to \infty} \Pr[|Y_1 - \lambda| \ge \epsilon] \\ &\ge \lim_{n \to \infty} \Pr[Y_1 \ge \epsilon + \lambda] \\ &\ge \lim_{n \to \infty} \Pr[Y_1 \ge 2\lambda] \\ &> 0. \end{align*}$$ Therefore, such an estimator is not consistent.