Exercises 47 and 48 refer to the following setting. Two independent random variables $X$and $Y$have the probability distributions, means, and standard deviations shown.

48. Difference Let the random variable $D=X-Y$.

(a) Find all possible values of D. Compute the probability that $D$takes each of these values. Summarize the probability distribution of$D$ in a table.
(b) Show that the mean of $D$is equal to${\mu }_X-{\mu }_Y$.
(c) Confirm that the variance of $D$is equal to ${\sigma }_X^2+{\sigma }_Y^2$.Find all possible values of $D$. Compute the probability that takes each of these values. Summarize the probability distribution of in a table.

Consider the random variable $T=X+Y$. The values of $\mathrm{T}$is to compute the probability that $T$takes each of these values for the probability distribution.

According to the information, the possible values of$D$ are:
$$\begin{gathered} 1-2=-1 \\ 1-4=-3 \\ 2-2=0 \\ 2-4=-2 \end{gathered}$$
Similarly,
$\begin{array}{r}5-2=3\\ 5-4=1\end{array}$
Determine the probabilities by:
$$\begin{gathered} P(D=-3)=P(X=1)\times P(Y=4) \\ =0.2(0.3) \\ =0.06 \end{gathered}$$
$$\begin{gathered} P(D=-2)=P(X=2)\times P(Y=4) \\ =0.5(0.3) \\ =0.15 \end{gathered}$$
$$\begin{gathered} P(D=-1)=P(X=1)\times P(Y=2) \\ =0.2(0.7) \\ =0.14 \end{gathered}$$
$$\begin{gathered} P(D=0)=P(X=2)\times P(Y=2) \\ =0.5(0.7) \\ =0.35 \end{gathered}$$
Also,
$$\begin{gathered} P(D=1)=P(X=5)\times P(Y=4) \\ =0.3(0.3) \\ =0.09 \end{gathered}$$
$$\begin{gathered} P(D=3)=P(X=5)\times P(Y=2) \\ =0.3(0.7) \\ =0.21 \end{gathered}$$

The random variable is $D=X-Y$. The mean of $D$is equal to ${\mu }_X-{\mu }_Y$.

The mean of$\mathrm{T}$ be calculated as:
$$\begin{gathered} \begin{array}{c}{\mu }_D=\sum x\times P\left(x\right)\end{array} \\ =-3\left(0.06\right)+(-2)\left(0.15\right)+\dots .+3\left(0.21\right) \\ =0.1 \end{gathered}$$

Then,

$$\begin{gathered} {\mu }_X-{\mu }_Y=2.7+2.6 \\ =0.1 \end{gathered}$$

Confirm the variance of $D$is equal to ${\sigma }_X^2+{\sigma }_Y^2$.

Determine the variance of D by:
$$\begin{gathered} \begin{array}{c}{\sigma }_D^2=\sqrt{\sum x^2\times P\left(x\right)-{\left(\sum x\times P\left(x\right)\right)}^2}\end{array} \\ =\sqrt{\left(-3^2\times 0.06+-2^2\times 0.15+\dots ..+3^2\times 0.21\right)-(0.1{)}^2} \\ =3.25 \end{gathered}$$

Then,

$$\begin{gathered} {\sigma }_X^2+{\sigma }_Y^2={1.55}^2+{0.917}^2 \\ =3.2433 \end{gathered}$$