#Transformations: Bivariate Random Variables

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pine flare
#

Hello, this exercise is about probability. I am currently studying transformations: Bivariate Random Variables. I am not sure about my results, as it seems suspicious to me that the variable (y_2) has disappeared. Thank you in advance.

\vspace{0.5cm}

\textbf{Problem:}

Let ( X_1 ) and ( X_2 ) be two random variables with a joint pmf given by

[
p_{X_1,X_2}(x_1, x_2) = c x_1 x_2 \quad \text{for} \quad x_1 = 1, 2, 3, 4, 5; \quad x_2 = 1, 2, 3, 4, 5; \quad \text{and zero elsewhere,}
]

where ( c ) is a constant that must be found. Consider the transformations ( Y_1 = 3X_1X_2 ) and ( Y_2 = X_2 ). Find the value of the product ( p_{Y_1,Y_2}(27, 3) \cdot p_{Y_1,Y_2}(18, 3) ).

\vspace{0.5cm}

\textbf{Solution:}

\textbf{Find c}
[
\sum_{x_1=1}^{5} \sum_{x_2=1}^{5} p_{X_1,X_2}(x_1, x_2) = 1
]

Let's calculate the sum:

[
\sum_{x_1=1}^{5} \sum_{x_2=1}^{5} c \cdot x_1 \cdot x_2 = c * 225 = 1
]

Then c = 1/225

\textbf{Find the values of ( w_1(y_1, y_2) ) and ( w_2(y_1, y_2) ):}

[
w_1(y_1, y_2) = \frac{y_1}{3y_2}
]
[
w_2(y_1, y_2) = y_2
]

\textbf{Find the support of Y}

[
Y = \left{ (y_1, y_2) : \begin{array}{l}
y_1 = 3, 6, 9, 12, 15, 18, 24, 27, 30, 36, 45, 48, 60, 75, \
y_2 = 1, 2, 3, 4, 5, \text{ with } y_2 \leq y_1
\end{array} \right}
]

\textbf{Sustitue (x_1) and (w_2)}

[
p_{Y_1,Y_2}(y_1, y_2) = \frac{1}{225} * \frac{y_1}{3*y_2} * y_2 = \frac{y_1}{675}
]

\textbf{Calculate (p_{Y_1,Y_2}(27, 3))}

[
p_{Y_1,Y_2}(27, 3) = \frac{27}{675} = \frac{1}{25}
]

\textbf{Calculate (p_{Y_1,Y_2}(18, 3))}

[
p_{Y_1,Y_2}(27, 3) = \frac{18}{675} = \frac{2}{75}
]

\textbf{Solution is (\frac{1}{25}*\frac{2}{75} = \frac{2}{1875})}

[
]

toxic birchBOT
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lyric hazelBOT
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@pine flare

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Don't worry!

final bronze