91Ó°ÊÓ

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:
$$\begin{gathered} P(D=-3)=P(X=1)×P(Y=4) \\ =0.2(0.3) \\ =0.06 \end{gathered}$$
$$\begin{gathered} P(D=-2)=P(X=2)×P(Y=4) \\ =0.5(0.3) \\ =0.15 \end{gathered}$$
$$\begin{gathered} P(D=-1)=P(X=1)×P(Y=2) \\ =0.2(0.7) \\ =0.14 \end{gathered}$$
$$\begin{gathered} P(D=0)=P(X=2)×P(Y=2) \\ =0.5(0.7) \\ =0.35 \end{gathered}$$
Also,
$$\begin{gathered} P(D=1)=P(X=5)×P(Y=4) \\ =0.3(0.3) \\ =0.09 \end{gathered}$$
$$\begin{gathered} P(D=3)=P(X=5)×P(Y=2) \\ =0.3(0.7) \\ =0.21 \end{gathered}$$

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

The mean of$\mathrm{T}$ be calculated as:
$$\begin{gathered} \begin{array}{c}{\mathrm{μ}}_D=∑x×P\left(x\right)\end{array} \\ =−3\left(0.06\right)+(−2)\left(0.15\right)+….+3\left(0.21\right) \\ =0.1 \end{gathered}$$

Then,

$$\begin{gathered} {\mathrm{μ}}_X−{\mathrm{μ}}_Y=2.7+2.6 \\ =0.1 \end{gathered}$$

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

Determine the variance of D by:
$$\begin{gathered} \begin{array}{c}{\mathrm{σ}}_D^2=\sqrt{∑x^2×P\left(x\right)−{\left(∑x×P\left(x\right)\right)}^2}\end{array} \\ =\sqrt{\left(−3^2×0.06+−2^2×0.15+…..+3^2×0.21\right)−(0.1{)}^2} \\ =3.25 \end{gathered}$$

Then,

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