a. [tex]X[/tex], [tex]N_1[/tex], and [tex]N_2[/tex] each have mean 0, and by linearity of expectation we have
[tex]E[R_1]=E[X+N_1]=E[X]+E[N_1]=0[/tex]
[tex]E[R_2]=E[X+N_2]=E[X]+E[N_2]=0[/tex]
b. By definition of correlation, we have
[tex]\mathrm{Corr}[R_1,R_2]=\dfrac{\mathrm{Cov}[R_1,R_2]}{{\sigma_{R_1}}{\sigma_{R_2}}}[/tex]
where [tex]\mathrm{Cov}[/tex] denotes the covariance,
[tex]\mathrm{Cov}[R_1,R_2]=E[(R_1-E[R_1])(R_2-E[R_2])][/tex]
[tex]=E[R_1R_2]-E[R_1]E[R_2][/tex]
[tex]=E[R_1R_2][/tex]
[tex]=E[(X+N_1)(X+N_2)][/tex]
[tex]=E[X^2]+E[N_1X]+E[XN_2]+E[N_1N_2][/tex]
Because [tex]X,N_1,N_2[/tex] are mutually independent, the expectation of their products distributes over the factors:
[tex]\mathrm{Cov}[R_1,R_2]=E[X^2]+E[N_1]E[X]+E[X]E[N_2]+E[N_1]E[N_2][/tex]
[tex]=E[X^2][/tex]
and recall that variance is given by
[tex]\mathrm{Var}[X]=E[(X-E[X])^2][/tex]
[tex]=E[X^2]-E[X]^2[/tex]
so that in this case, the second moment [tex]E[X^2][/tex] is exactly the variance of [tex]X[/tex],
[tex]\mathrm{Cov}[R_1,R_2]=E[X^2]={\sigma_X}^2[/tex]
We also have
[tex]{\sigma_{R_1}}^2=\mathrm{Var}[R_1]=\mathrm{Var}[X+N_1]=\mathrm{Var}[X]+\mathrm{Var}[N_1]={\sigma_X}^2+{\sigma_{N_1}}^2[/tex]
and similarly,
[tex]{\sigma_{R_2}}^2={\sigma_X}^2+{\sigma_{N_2}}^2[/tex]
So, the correlation is
[tex]\mathrm{Corr}[R_1,R_2]=\dfrac{{\sigma_X}^2}{\sqrt{\left({\sigma_X}^2+{\sigma_{N_1}}^2\right)\left({\sigma_X}^2+{\sigma_{N_2}}^2\right)}}[/tex]
c. The variance of [tex]R_1+R_2[/tex] is
[tex]{\sigma_{R_1+R_2}}^2=\mathrm{Var}[R_1+R_2][/tex]
[tex]=\mathrm{Var}[2X+N_1+N_2][/tex]
[tex]=4\mathrm{Var}[X]+\mathrm{Var}[N_1]+\mathrm{Var}[N_2][/tex]
[tex]=4{\sigma_X}^2+{\sigma_{N_1}}^2+{\sigma_{N_2}}^2[/tex]
