document.write( "Question 1192148: Suppose the probability density of X is given by
\n" ); document.write( "f(x) = kxe^(-x^2 ), X>0
\n" ); document.write( " = 0, otherwise\r
\n" ); document.write( "\n" ); document.write( "(a) Find the value of k.
\n" ); document.write( "(b) Find the distribution function of X, i.e., the cumulative density function of X.\r
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Algebra.Com's Answer #849033 by CPhill(1959)\"\" \"About 
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**a) Find the value of k**\r
\n" ); document.write( "\n" ); document.write( "* **Condition for a valid probability density function (PDF):**
\n" ); document.write( " * The total area under the PDF curve must be equal to 1.
\n" ); document.write( " * Mathematically, this means: ∫[from 0 to ∞] f(x) dx = 1\r
\n" ); document.write( "\n" ); document.write( "* **Set up the integral:**
\n" ); document.write( " * ∫[from 0 to ∞] kxe^(-x^2) dx = 1\r
\n" ); document.write( "\n" ); document.write( "* **Solve the integral:**
\n" ); document.write( " * Let u = -x^2
\n" ); document.write( " * du = -2x dx
\n" ); document.write( " * dx = -du / (2x)\r
\n" ); document.write( "\n" ); document.write( " * Substitute: ∫[from 0 to ∞] kxe^u * (-du / (2x))
\n" ); document.write( " = -k/2 ∫[from 0 to -∞] e^u du
\n" ); document.write( " = -k/2 [e^u] [from 0 to -∞]
\n" ); document.write( " = -k/2 [0 - 1]
\n" ); document.write( " = k/2\r
\n" ); document.write( "\n" ); document.write( "* **Solve for k:**
\n" ); document.write( " * k/2 = 1
\n" ); document.write( " * k = 2\r
\n" ); document.write( "\n" ); document.write( "**Therefore, the value of k is 2.**\r
\n" ); document.write( "\n" ); document.write( "**b) Find the distribution function of X (cumulative density function)**\r
\n" ); document.write( "\n" ); document.write( "* **Definition:** The cumulative distribution function (CDF), F(x), gives the probability that the random variable X takes on a value less than or equal to x.\r
\n" ); document.write( "\n" ); document.write( "* **Calculate F(x):**
\n" ); document.write( " * For x ≤ 0: F(x) = P(X ≤ x) = 0 (since f(x) = 0 for x ≤ 0)
\n" ); document.write( " * For x > 0:
\n" ); document.write( " * F(x) = P(X ≤ x) = ∫[from 0 to x] f(t) dt
\n" ); document.write( " = ∫[from 0 to x] 2te^(-t^2) dt \r
\n" ); document.write( "\n" ); document.write( " * Let u = -t^2
\n" ); document.write( " * du = -2t dt
\n" ); document.write( " * dt = -du / (2t)\r
\n" ); document.write( "\n" ); document.write( " * Substitute: ∫[from 0 to -x^2] 2te^u * (-du / (2t))
\n" ); document.write( " = -∫[from 0 to -x^2] e^u du
\n" ); document.write( " = -[e^u] [from 0 to -x^2]
\n" ); document.write( " = -(e^(-x^2) - e^0)
\n" ); document.write( " = 1 - e^(-x^2)\r
\n" ); document.write( "\n" ); document.write( "* **Combine the results:**\r
\n" ); document.write( "\n" ); document.write( " * F(x) =
\n" ); document.write( " * 0, for x ≤ 0
\n" ); document.write( " * 1 - e^(-x^2), for x > 0\r
\n" ); document.write( "\n" ); document.write( "**Therefore, the distribution function of X is:**\r
\n" ); document.write( "\n" ); document.write( " F(x) = {
\n" ); document.write( " 0, for x ≤ 0
\n" ); document.write( " 1 - e^(-x^2), for x > 0
\n" ); document.write( " }
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