Let X₁, X₂ be two independent normal random variables with means µ₁, µ₂ and standard deviations σ₁, σ₂, respectively. Consider Y = X₁ – X₂; µ₁=µ₂ =1, σ₁=1,σ₂= 2. Then,

To determine the distribution of the random variable $Y=X_1-X_2$, we use the properties of independent normal random variables.

We are given:
- $X_1$ is normally distributed with mean ${\mu }_1=1$ and standard deviation ${\sigma }_1=1$. Thus, the variance of $X_1$ is ${\sigma }_1^2=1^2=1$.
- $X_2$ is normally distributed with mean ${\mu }_2=1$ and standard deviation ${\sigma }_2=2$. Thus, the variance of $X_2$ is ${\sigma }_2^2=2^2=4$.
- $X_1$ and $X_2$ are independent.

Step 1: Determine the distribution type of Y
Any linear combination of independent normal random variables is also normally distributed. Since $Y=X_1-X_2$ is a linear combination of the independent normal random variables $X_1$ and $X_2$, $Y$ must be normally distributed.

Step 2: Calculate the mean of Y
Using the linearity of expectation:
$E\left[Y\right]=E[X_1-X_2]=E\left[X_1\right]-E\left[X_2\right]$
Substituting the given mean values:
$E\left[Y\right]=1-1=0$

Step 3: Calculate the variance of Y
Since $X_1$ and $X_2$ are independent, their covariance is 0. The variance of the difference is:
$\mathrm{Var}\left(Y\right)=\mathrm{Var}(X_1-X_2)=\mathrm{Var}\left(X_1\right)+\mathrm{Var}\left(X_2\right)$
Substituting the variance values:
$\mathrm{Var}\left(Y\right)={\sigma }_1^2+{\sigma }_2^2=1+4=5$

Therefore, $Y$ is normally distributed with mean 0 and variance 5.