Probability Density of Sum of Random Variables

In summary: Therefore, the density for Z = X + Y is e^{-z} and it integrates to 1. In summary, the density for Z is given by f_Z(z) = e^{-z}, which can be sketched as a decaying exponential curve and integrates to 1.
  • #1
snipez90
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Homework Statement


Suppose X and Y are independent random variables with X following a uniform distribution on (0,1) and Y exponentially distributed with parameter [itex]\lambda = 1[/itex]. Find the density for Z = X + Y. Sketch the density and verify it integrates to 1.



Homework Equations


If Z = X + Y, and X and Y are independent,

[tex]f_z(z) = \int_{-\infty}^{\infty}f_Y(z-x)f_X(x)\,dx[/tex]


The Attempt at a Solution


I am trying to catch up on stats. I think I screwed up somewhere in this problem:

If X ~ Unif(0,1), then [itex]f_X(x)[/itex] should just be 1 on (0,1) right? Also if the parameter of the exponential distribution is 1, then [itex]f_Y(z-x) = e^{-(z-x)}?[/itex]

But then the expression for [itex]f_z(z)[/itex] doesn't seem to make much sense. Note I have no idea what I am doing.
 
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  • #2



Thank you for posting this problem. Let's go through it step by step to find the solution.

First, you are correct that if X is uniformly distributed on (0,1), then the density function is just 1 on that interval. So we have f_X(x) = 1 for x ∈ (0,1).

Next, if Y is exponentially distributed with parameter λ = 1, then the density function is f_Y(y) = λe^{-λy} = e^{-y}.

Now, we want to find the density function for Z = X + Y. This means we need to find the distribution of Z, which is given by the convolution of X and Y. As you correctly stated, this is given by the integral:

f_Z(z) = \int_{-\infty}^{\infty}f_Y(z-x)f_X(x)\,dx

Substituting in our expressions for f_X(x) and f_Y(z-x), we have:

f_Z(z) = \int_{-\infty}^{\infty}e^{-(z-x)}\cdot 1\,dx

Now, we can simplify this integral by using the property that e^{-(z-x)} = e^{-z}e^x. This gives us:

f_Z(z) = e^{-z}\int_{-\infty}^{\infty}e^x\,dx

The integral on the right is just the integral of the exponential function, which evaluates to e^x. So we have:

f_Z(z) = e^{-z}\cdot e^x \biggr|_{-\infty}^{\infty} = e^{-z}(e^\infty - e^{-\infty})

Since e^\infty is infinity and e^{-\infty} is 0, this simplifies to:

f_Z(z) = e^{-z}(∞ - 0) = e^{-z}

This is the density function for Z.

In order to sketch this density, we can plot the function f_Z(z) = e^{-z} on the z-axis, with z ∈ (-∞,∞). This will give us a decaying exponential curve.

To verify that this density integrates to 1, we can take the integral:

∫_{-\infty}^{\infty
 

What is the "Probability Density of Sum of Random Variables"?

The "Probability Density of Sum of Random Variables" is a mathematical concept that describes the likelihood of obtaining a certain sum when multiple random variables are added together. It is used in probability theory to calculate the probability distribution of a sum of random variables.

How is the "Probability Density of Sum of Random Variables" calculated?

The "Probability Density of Sum of Random Variables" is calculated by convolving the probability density functions of the individual random variables. This involves multiplying the probability density function of one random variable by the probability density function of another and integrating the result over all possible values.

What is the difference between the "Probability Density of Sum of Random Variables" and the "Probability Distribution of Sum of Random Variables"?

The "Probability Density of Sum of Random Variables" refers to the function that describes the likelihood of obtaining a certain sum, while the "Probability Distribution of Sum of Random Variables" refers to the set of all possible outcomes and their corresponding probabilities. The former is a continuous function, while the latter is a discrete set of values.

What is the importance of the "Probability Density of Sum of Random Variables" in real-world applications?

The "Probability Density of Sum of Random Variables" is important in many fields, including statistics, physics, and engineering. It is commonly used to model and analyze complex systems where multiple random variables are involved, such as in financial markets, weather forecasting, and risk analysis.

Can the "Probability Density of Sum of Random Variables" be used to predict the exact outcome of a sum of random variables?

No, the "Probability Density of Sum of Random Variables" can only provide a probability distribution of possible outcomes. It cannot predict the exact outcome of a sum of random variables, as the actual result will depend on the specific values of the individual random variables.

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