Marginal Probability Functions Discrete Random Variables X And Y

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Hey there, probability enthusiasts! Today, we're diving deep into the fascinating world of discrete random variables and their marginal probability functions. We'll be tackling a specific problem that involves finding the marginal probability functions of two discrete random variables, X and Y, given their joint probability function. So, buckle up and let's get started!

Problem Statement: Decoding the Joint Probability Function

Let's set the stage. We're given two discrete random variables, X and Y, and their joint probability function (p.f.) is defined as:

f(x, y) = e^(-2) / (x! * (y - x)!)

where x can take values from 0 to y, and y can take values from 0, 1, and so on. Our mission, should we choose to accept it, is to find the marginal probability functions of X and Y. Sounds like a thrilling quest, right?

Cracking the Marginal p.f. of X: A Summation Saga

So, how do we find the marginal p.f. of X, denoted as f_X(x)? Well, the key is to sum the joint probability function over all possible values of Y, given a specific value of X. Think of it as collapsing the joint distribution along the Y-axis to get the distribution of X alone. Mathematically, it looks like this:

f_X(x) = Σ[from y=x to ∞] f(x, y)

Notice that the summation starts from y = x because x cannot be greater than y according to the problem's conditions. Now, let's plug in our joint probability function:

f_X(x) = Σ[from y=x to ∞] e^(-2) / (x! * (y - x)!)

Alright, now comes the fun part – simplifying this summation. We can pull out the e^(-2) and x! terms since they don't depend on y:

f_X(x) = (e^(-2) / x!) * Σ[from y=x to ∞] 1 / (y - x)!

To make things clearer, let's perform a change of variable. Let's say z = y - x. Then, when y = x, z = 0, and as y approaches infinity, z also approaches infinity. So, our summation transforms into:

f_X(x) = (e^(-2) / x!) * Σ[from z=0 to ∞] 1 / z!

Now, does that summation look familiar? It should! It's the Taylor series expansion of e^1 (or simply e). Remember that e = Σ[from z=0 to ∞] 1 / z! So, we can replace the summation with e:

f_X(x) = (e^(-2) / x!) * e

Simplifying further, we get:

f_X(x) = e^(-1) / x!

And there you have it! The marginal p.f. of X is f_X(x) = e^(-1) / x!, where x = 0, 1, 2, and so on. This, my friends, is a Poisson distribution with a mean (λ) of 1. Isn't it amazing how these things connect?

Unraveling the Marginal p.f. of Y: Another Summation Adventure

Now, let's turn our attention to finding the marginal p.f. of Y, denoted as f_Y(y). This time, we need to sum the joint probability function over all possible values of X, given a specific value of Y. So, we're essentially collapsing the joint distribution along the X-axis to get the distribution of Y alone. The formula looks like this:

f_Y(y) = Σ[from x=0 to y] f(x, y)

Here, the summation runs from x = 0 to x = y because x can only take values between 0 and y. Let's substitute our joint probability function:

f_Y(y) = Σ[from x=0 to y] e^(-2) / (x! * (y - x)!)

Again, let's try to simplify this summation. We can pull out the e^(-2) term since it doesn't depend on x:

f_Y(y) = e^(-2) * Σ[from x=0 to y] 1 / (x! * (y - x)!)

Now, this summation looks a bit trickier than the last one. To make it more manageable, let's multiply and divide by y!:

f_Y(y) = e^(-2) * (1/y!) * Σ[from x=0 to y] y! / (x! * (y - x)!)

Why did we do that? Because now, the term inside the summation, y! / (x! * (y - x)!), is the binomial coefficient, which we can write as "y choose x" or (y choose x). So, our equation becomes:

f_Y(y) = e^(-2) * (1/y!) * Σ[from x=0 to y] (y choose x)

Now, remember the binomial theorem? It states that (1 + 1)^y = Σ[from x=0 to y] (y choose x). And (1 + 1)^y is simply 2^y. This is a crucial step! So, we can replace the summation with 2^y:

f_Y(y) = e^(-2) * (1/y!) * 2^y

And that's it! The marginal p.f. of Y is f_Y(y) = e^(-2) * 2^y / y!, where y = 0, 1, 2, and so on. This, my friends, is another familiar distribution – a Poisson distribution, but this time with a mean (λ) of 2. How cool is that?

Key Takeaways: Summing it All Up

Let's recap what we've accomplished today:

  • We successfully found the marginal p.f. of X by summing the joint probability function over all possible values of Y. We discovered that X follows a Poisson distribution with a mean of 1.
  • We also successfully found the marginal p.f. of Y by summing the joint probability function over all possible values of X. We learned that Y follows a Poisson distribution with a mean of 2.
  • We utilized some handy mathematical tools like the Taylor series expansion of e and the binomial theorem to simplify our summations. These tools are your friends in the world of probability!

Conclusion: The Power of Marginal Distributions

Understanding marginal distributions is crucial in probability and statistics. They allow us to analyze the behavior of individual random variables within a joint distribution. By summing over the other variables, we essentially isolate the distribution of the variable we're interested in. This gives us valuable insights into the individual characteristics of each variable.

So, the next time you encounter a joint probability function, don't be intimidated! Remember the power of summation and the tools we've discussed today. You'll be able to unravel the marginal distributions and gain a deeper understanding of the underlying random variables. Keep exploring, keep learning, and keep having fun with probability!