We have two independent RV: $X_{1}$ and $X_{2}$ with respective characteristic functions $\varphi_{1}(t)=(1-3it)^{-1}$, $\varphi_{2}(t)=(1-\frac{it}{2})^{-2}$. We have to calculate covariance of variables $X_{1}+3X_{2}$ and $X_{1}^{2}$.
From characteristic functions we can deduce that $X_{1}$ has gamma distribution with $\lambda=1/3, s=1$ and $X_{2}$ has gamma with $\lambda=2,s=2$. Also because they are independent we know their joint distribution $f_{X_{1},X_{2}}(x_{1},x_{2})$.
And: $$cov(X_{1}+3X_{2},X_{1}^{2})=E[(X_{1}+3X_{2})X_{1}^{2}]-E[X_{1}+3X_{2}]E[X_{1}^{2}]$$ $$=E[X_{1}^{3}]+3E[X_{1}^{2}]E[X_{2}]-E[X_{1}]E[X_{1}^{2}]-3E[X_{2}]E[X_{1}^{2}]$$ $=E[X_{1}^{3}]-E[X_{1}]E[X_{1}^{2}]$
Is this correct way to solve this? Obviously we can calculate $E[X_{i}^{k}]$ from characteristic functions.
Yes, but I would go for: $$\text{Cov}(X_1+3X_2,X_1^2)=\text{Cov}(X_1,X_1^2)+3\text{Cov}(X_2,X_1^2)=\text{Cov}(X_1,X_1^2)=\mathbb EX_1^3-\mathbb EX_1\mathbb EX_1^2$$The independence explanes the second equality.