Question

let x_1, x_2, x_3 be a random variable from a normal distribution with mean does not equal to 0 and variance = 1/24. What are the values of a and b, respectively, in order for L=ax_1+4x_2+bx_3 to have a standard normal distribution?

Answer 1

standard normal distribution properties

Answer 2

mean 0 and var 1

Answer 3

how did you get that?

Answer 4

I thought var=1/24

Answer 5

we have to get that for standard normal distribution

Answer 6

You want the moment generating function for \(L\) to be the same as the MGF for the standard normal distribution, which means you require that
\[M_L(t)=M_{aX_1+4X_2+bX_3}(t)=M_{X_1}(at)M_{X_2}(4t)M_{X3}(bt)=\exp\left(\frac{t^2}{2}\right)\]For any normal distribution with mean \(\mu\) and standard deviation \(\sigma\), the corresponding MGF for any random variable \(X\) with such a distribution is
\[M_X(t)=\exp\left(\mu t+\sigma^2\frac{t^2}{2}\right)\]You're given that the mean and variance for each of \(X_i\) respectively is non-zero and \(\dfrac{1}{24}\), so you have
\[\begin{align*}
M_L(t)&=\exp\left(\mu_1(at)+\frac{(at)^2}{48}\right)\exp\left(\mu_2(4t)+\frac{t^2}{3}\right)\exp\left(\mu_3(bt)+\frac{(bt)^2}{48}\right)\\
&=\exp\left((a\mu_1+4\mu_2+b\mu_3)t+\left(\frac{a^2}{24}+\frac{2}{3}+\frac{b^2}{24}\right)\frac{t^2}{2}\right)
\end{align*}\]\(L\) has mean \(0\) and variance/standard deviation \(1\), so
\[\begin{cases}
a\mu_1+4\mu_2+b\mu_3=0\\[1ex]
\dfrac{a^2}{24}+\dfrac{2}{3}+\dfrac{b^2}{24}=1\iff a^2+b^2=8
\end{cases}\]