# Prove that MSE of $T_1$ is less than MSE of $T_2$

Supposing that $X_1, X_2, ..., X_n$ are i.i.d with distribution $X_i \sim Bin(m_i,\theta) i=1,...,n$ where $m_i$ are positive integers. Consider the estimators of $\theta$

$T_1(X) = \frac{\sum_{i=1}^{n}X_i}{\sum_{i=1}^{n}m_i}$ and $T_2(X) = 1/n \sum_{i=1}^{n} \frac{X_i}{m_i}$

Prove that $MSE(T_1) < MSE(T_2)$.

I have found that $MSE(T_1) = \frac{\theta(1-\theta)}{\sum_{i=1}^{n}m_i}$ and $MSE(T_2) = \frac{\theta(1-\theta)}{n \sum_{i=1}^{n}1/m_i}$, but I couldn't prove that $\frac{n \sum_{i=1}^{n}1/m_i}{\sum_{i=1}^{n}m_i}$ is less than 1.

• Welcome to Cross Validated! Please take a moment to view our tour. This looks like a self-study question. If that is the case, please tag your question appropriately and read the associated wiki. – Tavrock Apr 3 '17 at 17:47
• The expression for $MSE(T_2)$ looks incorrect. – HStamper Apr 3 '17 at 18:08
• Yes, You're right. Actually $MSE(T_2) = \frac{\theta(1-\theta)}{n} \sum_{i=1}^{n}1/m_i$ – Mayara Apr 3 '17 at 18:30
• Correcting the expression of $MSE(T_2)$ I could prove this using that $(\sum a_i b_i)^2 \le (\sum a_i^2)(\sum b_i^2)$. Thanks for all help! – Mayara Apr 3 '17 at 18:33

Notice that $m_i\geq 1$. So:
$$n\sum_{i=1}^n\frac{1}{m_i}\leq n\leq\sum_{i=1}^n m_i.$$