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You have three uniform random variables $X$,$Y$,$Z$ that follow $U(0,1)$. Either using calculus or Monte Carlo simulations (i.e. a ton of random samples on the computer)

a. What is the probability density function of $X + Y + Z$?

b. What is the cumulative probability function of $X + Y + Z$? Graph it.

This should be a pretty simple question but I am confused about "uniform random variable". For the density function, I just added the three together. So it is 3 for 0 to 1 and 0 for the rest.

And the cumulative function is just X+Y+Z from 0 to 3 in a linear increasing manner.

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    Please share some thoughts of your own. How did you approach the problem? What did you try? What did not work? What is your background? – 2017-02-14
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    What is the joint distribution of $X$, $Y$ and $Z$? – 2017-02-14
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    I assumed it is 3? – 2017-02-14
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    To validate your hunch, can the suspected answer be a valid probability density function? Is it normalized? – 2017-02-14
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    What do you mean by normalized? I see it as a piece wise function – 2017-02-14
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    I am suggesting for you to check if the answer you think might be true satisfies properties that a probability density function should have. One such property, comes from expressing the probability of certain outcome (which equals what?) as an integral of the density. – 2017-02-14

1 Answers 1

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The frequency distribution of the sum of two random variables is the convolution of the frequency distributions.

That is

let $W =X+Y$ and let $f(x)$ be the frequency distribution for $W$

$f(x) = \int_{-\infty}^{\infty} u(\lambda)u(x-\lambda) \ d\lambda$

$u(\lambda) = 1$ when $\lambda \in [0,1]$ and $0$ otherwise.

And I am going to assume that $X,Y,Z$ are independent variables.

Suppose $x<1$

$f(x) = \int_{0}^{x} 1 \ d\lambda = x$

suppose $x >1$

$f(x) = \int_{x-1}^{1} 1 \ d\lambda = 2-x$

$f(x) = \begin{cases} x & 0\le x \le 1\\2-x & 1< x \le 2\end{cases}$

let $\Omega = X+Y+Z = W+Z$

Let $g(x)$ be the frequency distribution for $\Omega$

$g(x) = \int_{-\infty}^{\infty} f(\lambda)u(x-\lambda) \ d\lambda$

Suppose $x \in [0,1]$

$g(x) = \int_{0}^{x} \lambda \ d\lambda = \frac 12 x^2$

Suppose $x \in [1,2]$

$g(x) =$$\int_{x-1}^{1} f(\lambda) \ d\lambda + \int_{1}^{x} f(\lambda) \ d\lambda\\ \int_{x-1}^{1} \lambda \ d\lambda + \int_{1}^{x} 2-\lambda \ d\lambda\\ \frac 12 \lambda^2|_{x-1}^1 + 2\lambda - \frac 12 \lambda^2|_1^x\\ 3x - x^2 + \frac 32 = \frac 34 - (x-\frac 32)^2$

Suppose $x \in [2,3]$

$g(x) = \int_{x-1}^{2} (2-\lambda) \ d\lambda = 2\lambda - \frac 12 \lambda^2|_{x-1}^1=2 - 2x +2+ \frac 12 (1-x)^2\\ \frac 92 - 3x + \frac 12x^2 = \frac 12 (3-x)^2 $

$g(x) = \begin{cases} \frac 12 x^2 & 0\le x\le 1\\\frac 34 - (x-\frac 32)^2& 1

For the CDF, $G(x) = \int_0^x g(x) dx$

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    Since $X$ and $Y$ are continuous random variables, $\mathbb{P}\left(X+Y = y\right)$ is identically zero, you meant to say on the first line, for $Z = X+Y$, the density $f_{Z}\left(z\right)$ is... Besides, the OP fails to states the joint distribution of $X$, $Y$ and $Z$, hence you must be making further assumptions which need to be clearly stated. – 2017-02-14
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    @Sasha Do you think that this is cleaner? – 2017-02-14
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    Let X be a standard uniform r.v. Let Y be the random variable defined on the same probability space such that $Y(\omega) =X(\omega)$. Then the random variable $X+Y$ has a different density than what you wrote, right? The joint distribution of $X, Y$ must be stated. – 2017-02-14