For the random variable $X$ with the given probability mass function below, find the mean and variance.
(i) $f(x)=\dfrac1{10}$ for $x=2,5$ and $f(x)=\dfrac15$ for $x=0,1,3,4$.
(ii) $f(x)=\dfrac{4-x}{6}$ for $x=1,2,3$.
(iii) $f(x)=2(x-1)$ for $1

Question

For the random variable $X$ with the given probability mass function below, find the mean and variance.
(i) $f(x)=\dfrac1{10}$ for $x=2,5$ and $f(x)=\dfrac15$ for $x=0,1,3,4$.
(ii) $f(x)=\dfrac{4-x}{6}$ for $x=1,2,3$.
(iii) $f(x)=2(x-1)$ for $1<x<2$, and $0$ otherwise.
(iv) $f(x)=\dfrac12e^{-x/2}$ for $x>0$, and $0$ otherwise.

Accepted Answer

> [!TLDR]
>
> Each part uses $E(X)=\sum xf(x)$ (or $\int xf(x)dx$) and $E(X^2)=\sum x^2f(x)$ (or $\int x^2f(x)dx$), then $V(X)=E(X^2)-(E(X))^2$.

**Part (i): $f(x)=	frac1{10}$ at $x=2,5$; $f(x)=	frac15$ at $x=0,1,3,4$.**
$E(X)=2\!\left(	frac1{10}ight)+5\!\left(	frac1{10}ight)+0\!\left(	frac15ight)+1\!\left(	frac15ight)+3\!\left(	frac15ight)+4\!\left(	frac15ight)=	frac7{10}+	frac85=	frac7{10}+	frac{16}{10}=	frac{23}{10}=2.3$.
$E(X^2)=4\!\left(	frac1{10}ight)+25\!\left(	frac1{10}ight)+0+1\!\left(	frac15ight)+9\!\left(	frac15ight)+16\!\left(	frac15ight)=	frac{29}{10}+	frac{26}5=	frac{29}{10}+	frac{52}{10}=	frac{81}{10}=8.1$.
$V(X)=8.1-2.3^2=8.1-5.29=2.81$.

**Part (ii): $f(x)=	frac{4-x}6$ at $x=1,2,3$, i.e. $f(1)=	frac12,f(2)=	frac13,f(3)=	frac16$.**
$E(X)=1\!\left(	frac12ight)+2\!\left(	frac13ight)+3\!\left(	frac16ight)=	frac12+	frac23+	frac12=1+	frac23=	frac53$.
$E(X^2)=1\!\left(	frac12ight)+4\!\left(	frac13ight)+9\!\left(	frac16ight)=	frac12+	frac43+	frac32=2+	frac43=	frac{10}3$.
$V(X)=	frac{10}3-\left(	frac53ight)^2=	frac{10}3-	frac{25}9=	frac{30}9-	frac{25}9=	frac59$.

**Part (iii): $f(x)=2(x-1)$ on $(1,2)$.**
$E(X)=\displaystyle\int_1^2 x\cdot2(x-1)\,dx=2\int_1^2(x^2-x)\,dx=2\left[\dfrac{x^3}3-\dfrac{x^2}2ight]_1^2=2\left(\dfrac23-\left(-\dfrac16ight)ight)=2\cdot\dfrac56=\dfrac53$.
$E(X^2)=\displaystyle\int_1^2 x^2\cdot2(x-1)\,dx=2\int_1^2(x^3-x^2)\,dx=2\left[\dfrac{x^4}4-\dfrac{x^3}3ight]_1^2=2\left(\dfrac43-\left(-\dfrac1{12}ight)ight)=2\cdot\dfrac{17}{12}=\dfrac{17}6$.
$V(X)=\dfrac{17}6-\left(\dfrac53ight)^2=\dfrac{17}6-\dfrac{25}9=\dfrac{51}{18}-\dfrac{50}{18}=\dfrac1{18}$.

**Part (iv): $f(x)=	frac12e^{-x/2}$ on $x>0$ (exponential, rate $\lambda=	frac12$).** Standard exponential results: $E(X)=\dfrac1\lambda=2$, $V(X)=\dfrac1{\lambda^2}=4$.
> [!ANSWER]
> (i) mean $=2.3$, variance $=2.81$. (ii) mean $=\dfrac53$, variance $=\dfrac59$. (iii) mean $=\dfrac53$, v

Q.For the random variable $X$ with the given probability mass function below, find the mean and variance.

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Q.For the random variable $X$ with the given probability mass function below, find the mean and variance.

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