Question

 If the random variable X takes the values 1, 2, 3 and 4 such that

P(X = 1) = 3P(X = 2) = P(X = 3) = 5P(X = 4), find the probability distribution and

cumulative distribution function of X.

Answer 1

Let $P(X=1)=p;$

$P(X=2)=\cfrac{p}{3};\\ P(X=3)=p;\\ P(X=4)=\cfrac{p}{5};\\ p+\cfrac{p}{3}+p+\cfrac{p}{5}=1;\\ 2p+\cfrac{5p+3p}{30}=1;\\ p=\cfrac{15}{38}.$

The probability distribution:

$P(X=1)=\cfrac{15}{38},P(X=2)=\cfrac{5}{38}, \\P(X=3)=\cfrac{15}{38},P(X=4)=\cfrac{3}{38}.$

If $x<1, F_X(x)=P(X\le x)=0.$

Next, if $1\le x<2, F_X(x)=P(X\le x)=P(X=1)=\cfrac{15}{38}.$

Next, if $2\le x<3,$

$F_X(x)=P(X\le x)=P(X=1)+P(X=2)=\cfrac{15}{38}+\cfrac{5}{38}=\cfrac{20}{38}.$

Next, if $3\le x<4,$

$F_X(x)=P(X\le x)=\\ =P(X=1)+P(X=2)+P(X=3)=\\ =\cfrac{15}{38}+\cfrac{5}{38}+\cfrac{15}{38}=\cfrac{35}{38}.$

Finally, if $x>4,$

$F_X(x)=P(X\le x)=\\ =P(X=1)+P(X=2)+P(X=3)+P(X=4)=\\ =\cfrac{15}{38}+\cfrac{5}{38}+\cfrac{15}{38}+\cfrac{3}{38}=\cfrac{38}{38}=1.$

We have the cumulative distribution function of X:

$F_X(x)=\begin{cases} 0 & & &\text{for } x<1 \\ \cfrac{15}{38} & & &\text{for } 1\le x<2\\ \cfrac{20}{38} & & &\text{for } 2\le x<3\\ \cfrac{35}{38} & & &\text{for } 3\le x<4\\ 1 & & &\text{for } x\ge4. \end{cases}$