- #1
Pietair
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Homework Statement
Show that the expectation operator E() is a linear operator, or, implying:
[tex]E(a\bar{x}+b\bar{y})=aE(\bar{x})+bE(\bar{y})[/tex]
Homework Equations
[tex]E(\bar{x})=\int_{-\infty}^{+\infty}xf_{\bar{x}}(x)dx[/tex]
With [tex]f_{\bar{x}}[/tex] the probability density function of random variable x.
The Attempt at a Solution
[tex]aE(\bar{x})=a\int_{-\infty}^{+\infty}xf_{\bar{x}}(x)dx[/tex] and:
[tex]bE(\bar{y})=b\int_{-\infty}^{+\infty}yf_{\bar{y}}(y)dy[/tex]
Introducing a new random variable:
[tex]\bar{v}=a\bar{x}+b\bar{y}[/tex]
Then:
[tex]E(\bar{v})=E(a\bar{x}+b\bar{y})=\int_{-\infty}^{+\infty}vf_{\bar{v}}(v)dv=\int_{-\infty}^{+\infty}(ax+by)f_{\bar{v}}(v)dv[/tex]
And accordingly:
[tex]E(a\bar{x}+b\bar{y})=\int_{-\infty}^{+\infty}(ax+by)f_{\bar{v}}(v)dv=a\int_{-\infty}^{+\infty}xf_{\bar{v}}(v)dv+b\int_{-\infty}^{+\infty}yf_{\bar{v}}(v)dv[/tex]
So what remains to proof is that:
[tex]a\int_{-\infty}^{+\infty}xf_{\bar{v}}(v)dv+b\int_{-\infty}^{+\infty}yf_{\bar{v}}(v)dv=a\int_{-\infty}^{+\infty}xf_{\bar{x}}(x)dx+b\int_{-\infty}^{+\infty}yf_{\bar{y}}(y)dy[/tex]
And now I am stuck... I don't know how I can relate the p.d.f. of random variable v to the p.d.f.'s of random variables x and y.
Thank you in advance!