Question

3. The joint PMF of discrete random variables $X$ and $Y$ is given by $$p_{X,Y}(x, y) = \begin{cases} k(2x + y), & x = 1, 2; y = 1, 2 \\ 0, & \text{elsewhere} \end{cases}$$ (a) Find $k$. (b) Find the joint CDF of $X$ and $Y$. (c) Find the marginal PMF of $X$ and $Y$. (d) Are $X$ and $Y$ independent random variables? (e) Find the conditional PMF $p_{X|Y}(x|y)$ for $y = 1, 2$.

          3. The joint PMF of discrete random variables $X$ and $Y$ is given by
$$p_{X,Y}(x, y) = \begin{cases} k(2x + y), & x = 1, 2; y = 1, 2 \\ 0, & \text{elsewhere} \end{cases}$$
(a) Find $k$.
(b) Find the joint CDF of $X$ and $Y$.
(c) Find the marginal PMF of $X$ and $Y$.
(d) Are $X$ and $Y$ independent random variables?
(e) Find the conditional PMF $p_{X|Y}(x|y)$ for $y = 1, 2$.

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3 the joint pmf of discrete random variables x and y is given by pxyx y begincases k2x y x 1 2 y 1 2 0 textelsewhere endcases a find k b find the joint cdf of x and y c find the marginal pm 01836

Added by Jaime P.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert David Nguyen

02/19/2022

Step 1

For a valid PMF, the sum of all probabilities must be equal to 1. The possible values for $(x, y)$ are $(1, 1), (1, 2), (2, 1), (2, 2)$. Calculate $p_{X,Y}(x, y)$ for each pair: $p_{X,Y}(1, 1) = k(2(1) + 1) = 3k$ $p_{X,Y}(1, 2) = k(2(1) + 2) = 4k$ $p_{X,Y}(2, 1) = Show more…

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3. The joint PMF of discrete random variables $X$ and $Y$ is given by $$p_{X,Y}(x, y) = \begin{cases} k(2x + y), & x = 1, 2; y = 1, 2 \\ 0, & \text{elsewhere} \end{cases}$$ (a) Find $k$. (b) Find the joint CDF of $X$ and $Y$. (c) Find the marginal PMF of $X$ and $Y$. (d) Are $X$ and $Y$ independent random variables? (e) Find the conditional PMF $p_{X|Y}(x|y)$ for $y = 1, 2$.

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Transcript

00:01 Hi, i'm david and i'm here to help me answer your question.

00:03 Now let me bring up your question here.

00:06 In this question here we're going to discuss about the drawn probability match function of the two random variables x and y.

00:14 Here we're given the function of the action y like this.

00:19 Now the question i ask me to find the volume of the k.

00:22 To do it, i will formally the table that will express the probability for any pair of the x and y.

00:28 So we have this one will be the x.

00:30 This will be the y.

00:32 X can have the value from 1 and 2, and y can have the value of the 1, 2, and 3.

00:48 Now here we have the probability function as a form the k times xi square times y.

00:57 So when x and y both equal to 1, and then we just have 1 times 1 times k which is equal to the k.

01:05 When x x x21, y, 2 we have equal to the 2k.

01:10 When x equal to 1, 1, 2 we get equal to the 3k.

01:14 Now when x equal to 2, y, 2 1 we have equal to the 4k, x equal 2, 2, so 4 times 2 will be 8k.

01:25 And then the last 1, 4 times 3 will be 12k.

01:30 Now i'm going to compute the marginal value.

01:33 So it will add up the row, we get equal to...

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