Probability density functionsThe probability density function (pdf) \(f(x)\) of a continuous random variable \(X\) is defined as the derivative of the cdf \(F(x)\): Show \[ f(x) = \dfrac{d}{dx}F(x). \] It is sometimes useful to consider the cdf \(F(x)\) in terms of the pdf \(f(x)\): \[ F(x) = \int_{-\infty}^{x} f(t)\;dt. \qquad\qquad(\ast) \] The pdf \(f(x)\) has two important properties:
The first property follows from the fact that the cdf \(F(x)\) is non-decreasing and \(f(x)\) is its derivative. The second property follows from equation (\(\ast\)) above, since \(F(x) \to 1\) as \(x \to \infty\), and so the total area under the graph of \(f(x)\) is equal to 1. An infinite variety of shapes are possible for a pdf, since the only requirements are the two properties above. The pdf may have one or several peaks, or no peaks at all; it may have discontinuities, be made up of combinations of functions, and so on. figure 5 shows a pdf with a single peak and some mild skewness. As is the case for a typical pdf, the value of the function approaches zero as \(x \to \infty\) and \(x \to -\infty\). Figure 5: A pdf may look something like this. We now explore how probabilities concerning the continuous random variable \(X\) relate to its pdf. The important result here is that \[ \Pr(a < X \leq b) = \int_a^b f(x)\;dx = \bigl[ F(x) \bigr]_a^b. \] This result follows from the fact that both sides are equal to \(F(b) - F(a)\). Notes.
Exercise 1Consider the function \[ f(x) = \begin{cases} 6x(1-x) &\text{if } 0 \leq x \leq 1, \\\\ 0 &\text{otherwise.} \end{cases} \]
Example: Random numbers, continuedConsider the continuous random variable \(U\) from the first random-number example. Then \(U \stackrel{\mathrm{d}}{=} \mathrm{U}(0,1)\). The pdf of \(U\) is given by \[ f_U(u) = \begin{cases} 1 &\text{if } 0 < u < 1, \\\\ 0 &\text{otherwise.} \end{cases} \] Figure 6: The probability density function of \(U \stackrel{\mathrm{d}}{=} \mathrm{U}(0,1)\). Exercise 2Consider the function \(f_V\) shown in figure 7; assume that \(f_V(v) = 0\) for \(v<0\) and \(v>1\). Figure 7: The probability density function of a random variable \(V\) with the triangular distribution.
The triangular pdf shown in figure 7 is the pdf of the average of two \(\mathrm{U}(0,1)\) random variables. That is, if \(U_1 \stackrel{\mathrm{d}}{=} \mathrm{U}(0,1)\) and \(U_2 \stackrel{\mathrm{d}}{=} \mathrm{U}(0,1)\) are independent, then \(V = \dfrac{1}{2}(U_1 + U_2)\) has the pdf in figure 7. This raises the question: What does the average of three independent \(\mathrm{U}(0,1)\) random variables look like? The answer is shown in figure 8. If \(U_i \stackrel{\mathrm{d}}{=} \mathrm{U}(0,1)\), for \(i=1,2,3\), and the three random variables are independent, then \(W = \dfrac{1}{3}(U_1 + U_2 + U_3)\) has the following pdf. Detailed description Figure 8: The probability density function of \(W\), the average of three independent \(\mathrm{U}(0,1)\) random variables. Next page - Content - Mean and variance of a continuous random variable What is the probability density function of a continuous random variable?The probability density function or PDF of a continuous random variable gives the relative likelihood of any outcome in a continuum occurring. Unlike the case of discrete random variables, for a continuous random variable any single outcome has probability zero of occurring.
How do you find the probability density function?We can differentiate the cumulative distribution function (cdf) to get the probability density function (pdf). This can be given by the formula f(x) = dF(x)dx d F ( x ) d x = F'(x). Here, f(x) is the pdf and F'(x) is the cdf.
How do you find the probability distribution of a continuous random variable?The probability distribution of a continuous random variable X is an assignment of probabilities to intervals of decimal numbers using a function f(x), called a density function, in the following way: the probability that X assumes a value in the interval [a,b] is equal to the area of the region that is bounded above ...
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