Respuesta :
Answer:
(a)
[tex]\begin{array}{cccc} {y} & {0} & {1} & {2} & {p(y|1)} &{0.2353}&{0.5882} & {0.1765}\ \end{array}[/tex]
(b)
[tex]\begin{array}{cccc} {y} & {0} & {1} & {2} & {p(y|2)} &{0.0962}&{0.2692} & {0.6346}\ \end{array}[/tex]
Step-by-step explanation:
Given
[tex]\begin{array}{ccccc}{} & {y} & { } & { }& { } & {x} & {0} & {1} & {2} & {total} & {0} &{0.1}&{0.03} & {0.01} & {0.14} & {1} &{0.08}&{0.20} & {0.06} &{0.34}& {2} &{0.05}&{0.14} & {0.33} &{0.52 } \ \end{array}[/tex]
Solving (a): Find PMF of y if x = 1
This implies that, we consider the dataset of the row where x = 1 i.e.
[tex]\begin{array}{ccccc}{} & {y} & { } & { }& { } & {x} & {0} & {1} & {2} & {total} & {1} &{0.08}&{0.20} & {0.06} &{0.34} \ \end{array}[/tex]
The PMF of y given x is calculated using:
[tex]Py|x(yi)=P(y=y_i|x)=\frac{P(y=x_i\ \&\ x)}{P(x)}[/tex]
When x = 1
[tex]P(x) = 0.34[/tex] --- the sum of the rows
So, we have:
[tex]Py|x(yi)=P(y=y_i|x)=\frac{P(y=x_i\ \&\ x)}{0.34}[/tex]
For i = 0 to 2, we have:
[tex]i = 0[/tex]
[tex]P(y=0|x)=\frac{P(y=0\ \&\ x = 1)}{0.34}[/tex]
[tex]P(y=0|x)=\frac{0.08}{0.34}[/tex]
[tex]P(y=0|x)=0.2353[/tex]
[tex]i =1[/tex]
[tex]P(y=1|x)=\frac{P(y=1\ \&\ x = 1)}{0.34}[/tex]
[tex]P(y=1|x)=\frac{0.20}{0.34}[/tex]
[tex]P(y=1|x)=0.5882[/tex]
[tex]i = 2[/tex]
[tex]P(y=2|x)=\frac{P(y=2\ \&\ x = 1)}{0.34}[/tex]
[tex]P(y=2|x)=\frac{0.06}{0.34}[/tex]
[tex]P(y=2|x)=0.1765[/tex]
So, the PMF of y given that x = 1 is:
[tex]\begin{array}{cccc} {y} & {0} & {1} & {2} & {p(y|1)} &{0.2353}&{0.5882} & {0.1765}\ \end{array}[/tex]
Solving (b): The interpretation of (b) is to find the PMF of y if x = 2
This implies that, we consider the dataset of the row where x = 2 i.e.
[tex]\begin{array}{ccccc}{} & {y} & { } & { }& { } & {x} & {0} & {1} & {2} & {total} & {2} &{0.05}&{0.14} & {0.33} &{0.52 }\ \end{array}[/tex]
When [tex]x = 2[/tex]
[tex]P(x) = 0.52[/tex]
So, we have:
[tex]Py|x(yi)=P(y=y_i|x)=\frac{P(y=x_i\ \&\ x)}{0.52}[/tex]
For i = 0 to 2, we have:
[tex]i = 0[/tex]
[tex]P(y=0|x)=\frac{P(y=0\ \&\ x = 2)}{0.52}[/tex]
[tex]P(y=0|x)=\frac{0.05}{0.52}[/tex]
[tex]P(y=0|x)=0.0962\\[/tex]
[tex]i =1[/tex]
[tex]P(y=1|x)=\frac{P(y=1\ \&\ x = 2)}{0.52}[/tex]
[tex]P(y=1|x)=\frac{0.14}{0.52}[/tex]
[tex]P(y=1|x)=0.2692[/tex]
[tex]i = 2[/tex]
[tex]P(y=2|x)=\frac{P(y=2\ \&\ x = 2)}{0.52}[/tex]
[tex]P(y=2|x)=\frac{0.33}{0.52}[/tex]
[tex]P(y=2|x)=0.6346[/tex]
[tex]P(y=0|x)=0.0962\\[/tex]
[tex]P(y=1|x)=0.2692[/tex]
[tex]P(y=2|x)=0.6346[/tex]
So, the PMF of y given that x = 1 is:
[tex]\begin{array}{cccc} {y} & {0} & {1} & {2} & {p(y|2)} &{0.0962}&{0.2692} & {0.6346}\ \end{array}[/tex]