$M_Y(t) = (e^{t_1}a_1 + e^{t_2}a_2 + \dots + e^{t_\ell}a_\ell)^n$ (multinomial distribution)
I know that
\begin{align} M_Y'(0) & = E(Y) \\[8pt] M_Y''(0) & = E(Y^2) \\ & \,\,\, \vdots \\ M_Y^k (0) & = E(Y^k) \end{align}
But I don't know how to do for two values $$E(Y_iY_j)$$
I think
$$E(Y_i Y_j) = \frac{dM_Y(0)}{dt_it_j}$$
But how would I calculate that ?
\begin{align} & M_{Y_1,\ldots,Y_\ell} (t_1,\ldots,t_\ell) \\[8pt] = {} & \operatorname E(e^{t_1Y_1+\cdots+t_\ell Y_\ell} ). \\[10pt] & \frac{\partial^2}{\partial t_1 \, \partial t_2} M_{Y_1,\ldots,Y_\ell} (t_1,\ldots,t_\ell) \\[10pt] = {} & \frac{\partial^2}{\partial t_1 \, \partial t_2} M_{Y_1,\ldots,Y_\ell} (t_1,\ldots,t_\ell) \\[10pt] = {} & \frac{\partial^2}{\partial t_1 \, \partial t_2} \operatorname E\left( e^{t_1 Y_1 + \cdots + t_\ell Y_\ell} \right) \\[10pt] = {} & \operatorname E\left( \frac{\partial^2}{\partial t_1 \, \partial t_2} e^{t_1Y_1+\cdots+t_\ell Y_\ell} \right) \\ & \text{(just why this step works} \\ & \phantom{(} \text{could bear examination.)} \\[10pt] = {} & \operatorname E \left( Y_1 Y_2 e^{t_1 Y_1 + \cdots + t_\ell Y_\ell} \right). \\[10pt] \text{Therefore } & \left. \frac{\partial^2}{\partial t_1\, \partial t_1} \right|_{(t_1,\ldots,t_\ell) \,=\, (0,\ldots,0)} M_{Y_1,\ldots,Y_\ell} (t_1,\ldots,t_\ell) \\[10pt] = {} & \operatorname E(Y_1 Y_2). \end{align}