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miss Akunina [59]
3 years ago
10

g SupposeXis a Gaussian random variable with mean 0 and varianceσ2X. SupposeN1is a Gaussian random variable with mean 0 and vari

anceσ21. SupposeN2is a Gaussianrandom variable with mean 0 and varianceσ22. AssumeX,N1,N2are all independentof each other. LetR1=X+N1R2=X+N2.(a) Find the mean ofR1andR2. That is findE[R1] andE[R2].(b) Find the correlationE[R1R2] betweenR1andR2.(c) Find the variance ofR1+R2.
Mathematics
1 answer:
Marysya12 [62]3 years ago
8 0

a. X, N_1, and N_2 each have mean 0, and by linearity of expectation we have

E[R_1]=E[X+N_1]=E[X]+E[N_1]=0

E[R_2]=E[X+N_2]=E[X]+E[N_2]=0

b. By definition of correlation, we have

\mathrm{Corr}[R_1,R_2]=\dfrac{\mathrm{Cov}[R_1,R_2]}{{\sigma_{R_1}}{\sigma_{R_2}}}

where \mathrm{Cov} denotes the covariance,

\mathrm{Cov}[R_1,R_2]=E[(R_1-E[R_1])(R_2-E[R_2])]

=E[R_1R_2]-E[R_1]E[R_2]

=E[R_1R_2]

=E[(X+N_1)(X+N_2)]

=E[X^2]+E[N_1X]+E[XN_2]+E[N_1N_2]

Because X,N_1,N_2 are mutually independent, the expectation of their products distributes over the factors:

\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]

=E[X^2]

and recall that variance is given by

\mathrm{Var}[X]=E[(X-E[X])^2]

=E[X^2]-E[X]^2

so that in this case, the second moment E[X^2] is exactly the variance of X,

\mathrm{Cov}[R_1,R_2]=E[X^2]={\sigma_X}^2

We also have

{\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

and similarly,

{\sigma_{R_2}}^2={\sigma_X}^2+{\sigma_{N_2}}^2

So, the correlation is

\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)}}

c. The variance of R_1+R_2 is

{\sigma_{R_1+R_2}}^2=\mathrm{Var}[R_1+R_2]

=\mathrm{Var}[2X+N_1+N_2]

=4\mathrm{Var}[X]+\mathrm{Var}[N_1]+\mathrm{Var}[N_2]

=4{\sigma_X}^2+{\sigma_{N_1}}^2+{\sigma_{N_2}}^2

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