Calculating E(XY) for Jointly Distributed Discrete Random Variables | Homework

In summary, the given problem involves calculating the expected value of the product of two jointly distributed discrete random variables, X and Y, with a given probability mass function. The solution involves separating the double sum into two single sums and using known formulas for calculating these sums, where the constant c must be between 0 and 1 for the formulas to hold. The maximum value of the function ec (1-c), which plays a key role in the solution, can be found by using calculus and is equal to 1 when c equals 0.
  • #1
kingwinner
1,270
0

Homework Statement


X and Y are joointly distributed discrete random variables with probability mass function
pX,Y(x,y)=(c/ec)(1-c)xcy/y!, x,y=0,1,2,..., 0<c<1
Find E(XY)


Homework Equations


The Attempt at a Solution


Code:
By definition,
E(X[SUP]Y[/SUP])
  ∞    ∞
= ∑    ∑  x[SUP]y[/SUP] (c/e[SUP]c[/SUP])(1-c)[SUP]x[/SUP]c[SUP]y[/SUP]/y!
 x=0  y=0
How can we calculate this double sum? We can pull out the constant c/ec, but what's next?

Thanks for any help!
 
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  • #2
Do the y-sum first and use
[tex]
\sum_{y=0}^\infty{\frac{Z^y}{y!}}=e^Z
[/tex]
What is Z here?
Next do the x-sum using
[tex]
\sum_{x=0}^\infty{a^x}=\frac{1}{1-a}
[/tex]
What is a here? Is |a|<1?
 
  • #3
Pere Callahan said:
Do the y-sum first and use
[tex]
\sum_{y=0}^\infty{\frac{Z^y}{y!}}=e^Z
[/tex]
What is Z here?
Next do the x-sum using
[tex]
\sum_{x=0}^\infty{a^x}=\frac{1}{1-a}
[/tex]
What is a here? Is |a|<1?

The xy is giving me some trouble doing the y-sum, it depends on both x and y, and I cannot separate it into a product, what should I do?
 
  • #4
When doing the y-sum treat x as a constant.
 
  • #5
OK, so
E(XY)

= ∑ [ec (1-c)]x (c/ec)
x=0
={1/[1-ec (1-c)]} (c/ec)

Did I get it right?


Also, why is |ec (1-c)|<1 ? How can we see this? I can only show 0<c<1 => 0<1-c<1 => 0<(1-c)ec<ec < e, but we need (1-c)ec<ec < 1, how can we prove this? This is giving me so much headache...please help...


Thank you!
 
Last edited:
  • #6
Just out of curiousity, in this case, is it possible to sum over x FIRST, and then sum over y?

Code:
∞    ∞
∑    ∑  x[SUP]y[/SUP] (c/e[SUP]c[/SUP])(1-c)[SUP]x[/SUP]c[SUP]y[/SUP]/y!
y=0 x=0
 
  • #7
kingwinner said:
OK, so
E(XY)

= ∑ [ec (1-c)]x (c/ec)
x=0
={1/[1-ec (1-c)]} (c/ec)

Did I get it right?
Yes.
kingwinner said:
Also, why is |ec (1-c)|<1 ? How can we see this? I can only show 0<c<1 => 0<1-c<1 => 0<(1-c)ec<ec < e, but we need (1-c)ec<ec < 1, how can we prove this? This is giving me so much headache...please help...
Try and find the maximum of ec (1-c) on the interval [0,1]. You surely know how to find the maximum of a differentiable function:smile: You will find that the function has its maximum when c=0 and this maximum value is 1.

For your other question: Sure you can but it's a lot more complicated.
 
  • #8
"Try and find the maximum of ec (1-c) on the interval [0,1]. You surely know how to find the maximum of a differentiable function You will find that the function has its maximum when c=0 and this maximum value is 1."

OK, I got it! Thanks!
Is it possible to prove it without using calculus?
 
  • #9
kingwinner said:
Is it possible to prove it without using calculus?

Still wondering...
 

1. What is Expected Value?

The expected value, denoted as E(X), is a measure of the central tendency or average of a random variable X. It represents the theoretical long-term average of the outcomes of a random experiment.

2. How is Expected Value calculated?

The expected value of a discrete random variable X is calculated by multiplying each possible outcome by its respective probability and then summing up all the products. For a continuous random variable, the expected value is calculated by integrating the outcome values with respect to the probability density function.

3. What is the significance of Expected Value?

Expected value is an important concept in probability and statistics as it provides a single value that summarizes the outcomes of a random variable. It allows for comparisons between different experiments and serves as a basis for decision making in situations involving uncertainty.

4. What is the difference between Expected Value and Mean?

The expected value and mean are often used interchangeably, but they are not exactly the same. The mean is a measure of the central tendency of a data set, while the expected value is a theoretical concept used in probability and statistics to describe the average outcome of a random variable.

5. How is Expected Value used in decision making?

Expected value is used in decision making to assess the potential outcomes and risks associated with different options. By calculating the expected value of each option, one can make an informed decision and choose the option with the highest expected value, which is considered the most optimal choice.

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