In this question I tried the approach as
since,
E[Z]=0
E[Z]=a*E[X]+b
E[X]= -b/a
V[Z]=1
V(Z)=a^2*V(X)
V(X)=1/a^2
I am stuck after this.
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$\begingroup$ reference for maths typesetting on this site. $\endgroup$– Siong Thye GohCommented Oct 30, 2018 at 8:08
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1 Answer
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Solving for $a$,$$a=\pm\frac1{\sqrt{V(X)}}$$
then substitute to the first condition that you found. $$b = -aE[X]=\mp \frac{E[X]}{\sqrt{V(X)}}$$
I think the purpose of the exerise is to construct that
$$\pm\frac{X-E(X)}{\sqrt{V(X)}}$$ has mean $0$ and variance $1$.
answered Oct 30, 2018 at 8:04
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$\begingroup$ but I am still unable to find the values of a and b... first i thought that a=1 and b=0, but that's also not correct. $\endgroup$ Commented Oct 30, 2018 at 16:09
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$\begingroup$ erm.... I have given you the answer, $a = \pm \frac1{\sqrt{V(X)}}$ $\endgroup$ Commented Oct 30, 2018 at 16:21
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$\begingroup$ yeaa... but according to this answer isin't correct... and also I don't know the correct answer. $\endgroup$ Commented Oct 30, 2018 at 17:13
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$\begingroup$ what is the source of the question? $\endgroup$ Commented Oct 30, 2018 at 17:14
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$\begingroup$ Its a question from a random exercise i got through on internet. I have 4 attempts to solve this and i have used 3 attempts $\endgroup$ Commented Oct 31, 2018 at 6:59