Question

Determine the value of c so that the following functions represent the joint pmf of
the random variables X and Y .
i) f (x, y) = c(x + y +1), x = 0, 1, 2, 3 and y = 0, 1, 2 .
ii) f (x, y) = c(x^2 + y^2), x = −1, 1 and y = −2, 2

Answer 1

Answer on Question #64974 – Math – Statistics and Probability

Question

Determine the value of $c$ so that the following functions represent the joint pmf of the random variables $X$ and $Y$.

i) $f(x,y) = c(x + y + 1), x = 0,1,2,3; y = 0,1,2.$

Solution

Since $f(x, y)$ is a joint probability mass function (abbreviated p. m. f.) [1] then

$$\sum_{x=0}^{3} \sum_{y=0}^{2} c(x + y + 1) = 1$$

Now we shall expand this double sum [2].

Expanding the second sum we get:

$$c \sum_{x=0}^{3} (x + 0 + 1 + x + 1 + 1 + x + 2 + 1) = c \sum_{x=0}^{3} (3x + 6) = 1.$$

Expanding the first sum we get

$$c(3 \cdot 0 + 6 + 3 \cdot 1 + 6 + 3 \cdot 2 + 6 + 3 \cdot 3 + 6) = 42c = 1 \Rightarrow c = \frac{1}{42}.$$

Answer: $\frac{1}{42}$.

Question

Determine the value of $c$ so that the following functions represent the joint pmf of the random variables $X$ and $Y$.

ii) $f(x,y) = c(x^2 + y^2), x = -1,1; y = -2,2.$

Solution

We have the following distribution:

Since $f(x, y)$ is a p. m. f. then we have

$$5c + 5c + 5c + 5c = 1 \Rightarrow 20c = 1 \Rightarrow c = \frac{1}{20}.$$

Answer: $\frac{1}{20}$.

Answer provided by https://www.AssignmentExpert.com

References:

[1] PennState Eberly College of Science. STAT 414 Intro Probability Theory. Lesson 17. Two Discrete Random Variables. Retrieved from https://onlinecourses.science.psu.edu/stat414/node/104.

[2] Double Series. Retrieved from http://mathworld.wolfram.com/DoubleSeries.html.