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.

47. Sum Let the random variable$T=X+Y$.
(a) Find all possible values of $\mathrm{T}$. Compute the probability that $T$takes each of these values. Summarize the probability distribution of$\mathrm{T}$ in a table.
(b) Show that the mean of $T$is equal to${\mu }_{\mathrm{X}}+{\mu }_{\mathrm{Y}}$.
(c) Confirm that the variance of T is equal to ${\sigma }_X^2+{\sigma }_{\mathrm{Y}}^2$. Show that ${\sigma }_{\mathrm{T}}\ne {\sigma }_{\mathrm{X}}+{\sigma }_{\mathrm{Y}}$.

The random variable is $T=X+Y$. All possible values of $\mathrm{T}$. And also compute the probability that $T$takes each of the values for the probability distribution.

The possible values of$T$ are:
$$\begin{gathered} 1+2=3 \\ 1+4=5 \\ 2+2=4 \\ 2+4=6 \end{gathered}$$

Similarly,

$\begin{array}{r}5+2=7\\ 5+4=9\end{array}$

The probabilities as:

$$\begin{gathered} P(T=3)=P(X=1)\times P(Y=2) \\ =0.2(0.7) \\ =0.14 \end{gathered}$$

$$\begin{gathered} P(T=4)=P(X=2)\times P(Y=2) \\ =0.5(0.7) \\ =0.35 \end{gathered}$$

$$\begin{gathered} P(T=5)=P(X=1)\times P(Y=4) \\ =0.2(0.3) \\ =0.06 \end{gathered}$$

$$\begin{gathered} P(T=6)=P(X=2)\times P(Y=4) \\ =0.5(0.3) \\ =0.15 \end{gathered}$$

Also,

$$\begin{gathered} P(T=7)=P(X=5)\times P(Y=2) \\ =0.3(0.7) \\ =0.21 \end{gathered}$$

$$\begin{gathered} P(T=9)=P(X=5)\times P(Y=4) \\ =0.3(0.3) \\ =0.09 \end{gathered}$$

Two independent random variables $X$and $Y$have the probability distributions, means, and standard deviations. The mean of $T$is equal to ${\mu }_{\mathrm{X}}+{\mu }_{\mathrm{Y}}$.

The mean of$\mathrm{T}$ be determined as:
$$\begin{gathered} {\mu }_T=\sum x\times P\left(x\right) \\ =3\left(0.14\right)+4\left(0.35\right)+5\left(0.06\right)+6\left(0.15\right)+7\left(0.21\right)+9\left(0.09\right) \\ =5.3 \end{gathered}$$

Then,

$$\begin{gathered} {\mu }_X+{\mu }_Y=2.7+2.6 \\ =5.3 \end{gathered}$$

The variance of $T$is equal to ${\sigma }_X^2+{\sigma }_{\mathrm{Y}}^2$. And also show that ${\sigma }_{\mathrm{T}}\ne {\sigma }_{\mathrm{X}}+{\sigma }_{\mathrm{Y}}$.

To determine the variance of$\mathrm{T}$ as:
$$\begin{gathered} \begin{array}{c}{\sigma }_T=\sqrt{\sum x^2\times P\left(x\right)-{\left(\sum x\times P\left(x\right)\right)}^2}\end{array} \\ =\sqrt{\left(3^2\times 0.14+4^2\times 0.35+\dots \dots ...+9^2\times 0.09\right)-(5.3{)}^2} \\ =3.25 \end{gathered}$$

Then,

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

It could be shown as:

$$\begin{gathered} \begin{array}{c}{\sigma }_T={\sigma }_X+{\sigma }_Y\end{array} \\ =1.55+0.917 \\ =2.467 \end{gathered}$$