Continuous Random Variables: PDFs, CDFs, Expected Value, and Variance

Continuous Random Variables and Probability Density Functions (PDFs)

• Definition of a Continuous Random Variable: A random variable that can take on any value within a given range in the real number system, often an interval $[a, b]$ (where $a$ or $b$ can be infinite).

• Probability Density Function (PDF), denoted $f_X(x)$:

  • It's analogous to density in physics (e.g., mass density). If you integrate the density of an object, you get its total mass. In probability, integrating a PDF gives the total probability.

  • Properties:

    • The PDF must be non-negative: $f_X(x) \ge 0$ for all $x$.

    • The total area under the curve of the PDF over its entire range of possible values must equal one: $\int<em>{-\infty}^{\infty} f</em>X(x) dx = 1$.

    • For a defined range $[a, b]$, this becomes $\int<em>a^b f</em>X(x) dx = 1$.

  • Computing Probability: To find the probability that a continuous random variable $X$ falls within a certain subset (or interval) $A$ of its range, you integrate the PDF over that subset:
    $P(X \in A) = \int<em>A f</em>X(x) dx$
    For an interval $[c, d]$ within the range: $P(c \le X \le d) = \int<em>c^d f</em>X(x) dx$

Cumulative Distribution Function (CDF)

• Definition, denoted $F_X(x)$: The cumulative distribution function gives the probability that the random variable $X$ takes on a value less than or equal to a specific value $x$.

  • It represents the area under the PDF curve up to a point $x$.

  • Formula: $F<em>X(x) = P(X \le x) = \int</em>{-\infty}^{x} f_X(t) dt$
    (A dummy variable $t$ is often used for the integration to avoid confusion with the upper limit $x$).

• Relationship between PDF and CDF: By the Fundamental Theorem of Calculus, the derivative of the CDF gives the PDF:
  $f<em>X(x) = \frac{d}{dx} F</em>X(x)$

Expected Value and Variance for Continuous Random Variables

• Expected Value (Mean), denoted $E[X]$ or $\mu_X$:

  • It represents the long-term average value of the random variable.

  • Formula: $E[X] = \int<em>{-\infty}^{\infty} x \cdot f</em>X(x) dx$
    This is a weighted average where each value of $x$ is weighted by its probability density.

• Expected Value of a Function of X, denoted $E[g(X)]$:

  • For any function $g(x)$, its expected value is:
    $E[g(X)] = \int<em>{-\infty}^{\infty} g(x) \cdot f</em>X(x) dx$

• Variance, denoted $Var(X)$ or $\sigma_X^2$:

  • Measures the spread or dispersion of the data around the mean.

  • Formula: $Var(X) = E[(X - \mu<em>X)^2] = \int</em>{-\infty}^{\infty} (x - \mu<em>X)^2 f</em>X(x) dx$

  • Alternative and often easier formula: $Var(X) = E[X^2] - (E[X])^2$
    Where $E[X^2] = \int<em>{-\infty}^{\infty} x^2 f</em>X(x) dx$.

Example: Finding Constant 'c', Probability, Expected Value, and Variance

Let the PDF be $f_X(x) = c \cdot x^2$ for $x \in [0, 3]$ and $0$ otherwise.

1. Find the constant 'c'

• Condition: The total probability over the range must be 1.
  $\int_0^3 c x^2 dx = 1$

• Integration:
  $c \left[ \frac{x^3}{3} \right]_0^3 = 1$
  $c \left( \frac{3^3}{3} - \frac{0^3}{3} \right) = 1$
  $c \left( \frac{27}{3} \right) = 1$
  $c \cdot 9 = 1$
  $c = \frac{1}{9}$

• Result: The PDF is $f_X(x) = \frac{1}{9} x^2$ for $x \in [0, 3]$ and $0$ otherwise.

  • Visualization: This function represents the right half of a parabola between 0 and 3. The total area under this curve is 1.

2. Compute the probability $P(1 \le X \le 2)$

• Formula: $\int<em>1^2 f</em>X(x) dx$

• Calculation:
  $P(1 \le X \le 2) = \int<em>1^2 \frac{1}{9} x^2 dx$ $= \frac{1}{9} \left[ \frac{x^3}{3} \right]</em>1^2$
  $= \frac{1}{9} \left( \frac{2^3}{3} - \frac{1^3}{3} \right)$
  $= \frac{1}{9} \left( \frac{8}{3} - \frac{1}{3} \right)$
  $= \frac{1}{9} \left( \frac{7}{3} \right) = \frac{7}{27}$

3. Compute the Expected Value $E[X]$

• Formula: $E[X] = \int<em>0^3 x \cdot f</em>X(x) dx$

• Calculation:
  $E[X] = \int<em>0^3 x \cdot \frac{1}{9} x^2 dx = \int</em>0^3 \frac{1}{9} x^3 dx$
  $= \frac{1}{9} \left[ \frac{x^4}{4} \right]_0^3$
  $= \frac{1}{9} \left( \frac{3^4}{4} - \frac{0^4}{4} \right)$
  $= \frac{1}{9} \left( \frac{81}{4} \right) = \frac{9}{4}$

4. Compute the Variance $Var(X)$

• First, calculate $E[X^2]$:
  $E[X^2] = \int<em>0^3 x^2 \cdot f</em>X(x) dx = \int<em>0^3 x^2 \cdot \frac{1}{9} x^2 dx = \int</em>0^3 \frac{1}{9} x^4 dx$
  $= \frac{1}{9} \left[ \frac{x^5}{5} \right]_0^3$
  $= \frac{1}{9} \left( \frac{3^5}{5} - \frac{0^5}{5} \right)$
  $= \frac{1}{9} \left( \frac{243}{5} \right) = \frac{27}{5}$

• Now, calculate $Var(X)$: $Var(X) = E[X^2] - (E[X])^2$ Var(X) = \frac{27}{5} - \left( \frac{9}{4} ight)^2 $Var(X) = \frac{27}{5} - \frac{81}{16}$ $Var(X) = \frac{27 \cdot 16 - 81 \cdot 5}{80} = \frac{432 - 405}{80} = \frac{27}{80}$

  • The variance indicates the spread of the data. A higher variance means greater spread.

Example: U-Substitution in Finding 'c'

Let $f_X(x) = c x \cos(x^2)$ for $x \in [0, \sqrt{\pi/2}]$ and $0$ otherwise.

Finding the constant 'c'

• Condition: $\int_0^{\sqrt{\pi/2}} c x \cos(x^2) dx = 1$

• U-Substitution: Let $u = x^2$. Then $du = 2x dx$, so $x dx = \frac{1}{2} du$.

• Change Limits of Integration:

  • If $x=0$, then $u=0^2=0$.

  • If $x=\sqrt{\pi/2}$, then $u=(\sqrt{\pi/2})^2=\pi/2$.

• Substitute and Integrate:
  $\int<em>0^{\pi/2} c \cos(u) \frac{1}{2} du = 1$ $\frac{c}{2} \int</em>0^{\pi/2} \cos(u) du = 1$
  $\frac{c}{2} \left[ \sin(u) \right]_0^{\pi/2} = 1$
  $\frac{c}{2} (\sin(\pi/2) - \sin(0)) = 1$
  $\frac{c}{2} (1 - 0) = 1$
  $\frac{c}{2} = 1$
  $c = 2$

• Result: The PDF is $f_X(x) = 2x \cos(x^2)$ for $x \in [0, \sqrt{\pi/2}]$ and $0$ otherwise.

  • Note on Positivity: It's important to ensure the function is non-negative over its range. For $x \in [0, \sqrt{\pi/2}]$, $x$ is positive, and $x^2 \in [0, \pi/2]$. In this range, $\cos(x^2)$ is positive, so the entire PDF is positive.

Attempting Expected Value $E[X]$ for the above PDF

• Formula: $E[X] = \int<em>0^{\sqrt{\pi/2}} x \cdot (2x \cos(x^2)) dx = \int</em>0^{\sqrt{\pi/2}} 2x^2 \cos(x^2) dx$

• This integral is complex and cannot be solved by simple u-substitution or elementary integration by parts directly as presented. It may require advanced techniques like power series expansion (Taylor series) and term-by-term integration for approximation, or might be a type of Fresnel integral. This demonstrates that not all integrals for expected values are straightforward to compute analytically.