Question

If X₁ and X₂ are independent and identically distributed normal random variables with mean m and variance s², find the probability distribution function for U=X1-3X2/2.

Answer 1

Answer:-

Given that:-

If $X_{1}$ and $X_{2}$ are independent and identically distributed normal random variables with mean m and variance s2, find the probability distribution function for $U=X_{1}-3X_{2}/2$

Let $X_{1}$ & $X_{2}$ on independet & identically distribution r.v.s with mean $\mu$ &$\sigma ^{2}$

$X \sim N(\mu,\sigma ^{2})$

we know that

$M_{X_{1}}(t)=M_{X_{2}}(t)=e^{\mu t+\frac{t^{2}\sigma ^{2}}{2}}$

Let $U=\frac{X_{1}}{2}-\frac{3X_{2}}{2}$

mgf of U

$M_{U}(t)=\frac{MX_{1}}{2}-\frac{3X_{2}}{2}(t)$

$=\frac{MX_{1}}{2}(t),\frac{3X_{2}}{2}(-t)$   $\begin{Bmatrix} iid \end{Bmatrix}$

$=MX_{1}(\frac{t}{2}),MX_{2}(\frac{-3t}{2})$

$=e^{\frac{\mu t}{2}+\frac{t^{2}}{2}\frac{\sigma ^{2}}{4}},e^{\frac{-3\mu t}{2}+\frac{9 t^{2}}{4\times 2}\sigma ^{2}}$

$=e^{-\mu t+\frac{t^{2}}{2}(\frac{5 }{2}\sigma ^{2})}$

Which is mgf  $N(-\mu,\frac{5}{2}\sigma ^{2})$

Hence y uniduchess theorem

$U \sim N(-\mu,\frac{5}{2}\sigma ^{2})$