Question

The following is a p.d.f. (Probability density function) of a continuous random variable \[X\]:
\[f\left( x \right) = \dfrac{x}{{32}}{\rm{ }}0 < x < 8; = 0\] otherwise
Find the value of the c.d.f. at \[x = 0.5\] and \[x = 9\].

Answer 1

Hint: Here, we will use to find the formula of c.d.f. of a probability density function with the limits 0 and \[0.5\] to find the value of the c.d.f. at \[x = 0.5\]. Then, we will use to find the formula of c.d.f. of a probability density function with the limits 0 and 9. We will then use the property of definite integrals to split the integral into a sum of two integrals (with limits 0 to 8, and 8 to 9 respectively). Finally, we will integrate the expressions to find the value of the c.d.f. at \[x = 9\].
Formula Used: We will use the following formulas:
1.The c.d.f. of a p.d.f. \[f\left( x \right)\] is given as \[F\left( x \right) = \int\limits_{ - \infty }^x {f\left( x \right)} dx\].
2.The integral of a function of the form \[af\left( x \right)\] can be written as \[\int {af\left( x \right)} dx = a\int {f\left( x \right)} dx\].
3.The integral of a function of the form \[{x^n}\] can be written as \[\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C\].
4.The definite integral \[\int\limits_a^b {f\left( x \right)} dx\] can be written as the sum of the definite integrals \[\int\limits_a^c {f\left( x \right)} dx\] and \[\int\limits_c^b {f\left( x \right)} dx\], where \[a < c < b\].

Complete step-by-step answer:
We need to find the value of the cumulative density function at \[x = 0.5\].
The value \[x = 0.5\] lies between 0 and 8.
Thus, the lower limit becomes 0 and the upper limit becomes \[0.5\].
Therefore, substituting \[x = 0.5\] and \[f\left( x \right) = \dfrac{x}{{32}}\] in the c.d.f. \[F\left( x \right) = \int\limits_{ - \infty }^x {f\left( x \right)} dx\], we get
\[ \Rightarrow F\left( {0.5} \right) = \int\limits_0^{0.5} {\dfrac{x}{{32}}} dx\]
Now, we will integrate the expression.
The integral of a function of the form \[af\left( x \right)\] can be written as \[\int {af\left( x \right)} dx = a\int {f\left( x \right)} dx\].
Therefore, we get
\[ \Rightarrow F\left( {0.5} \right) = \dfrac{1}{{32}}\int\limits_0^{0.5} x dx\]
Rewriting the expression, we get
\[ \Rightarrow F\left( {0.5} \right) = \dfrac{1}{{32}}\int\limits_0^{0.5} {{x^1}} dx\]
The integral of a function of the form \[{x^n}\] can be written as \[\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C\].
Therefore, we get
\[ \Rightarrow F\left( {0.5} \right) = \dfrac{1}{{32}}\left. {\left( {\dfrac{{{x^{1 + 1}}}}{{1 + 1}}} \right)} \right|_0^{0.5}\]
Simplifying the expression, we get
\[ \Rightarrow F\left( {0.5} \right) = \dfrac{1}{{32}}\left. {\left( {\dfrac{{{x^2}}}{2}} \right)} \right|_0^{0.5}\]
Substituting the limits into the expression, we get
\[ \Rightarrow F\left( {0.5} \right) = \dfrac{1}{{32}}\left[ {\dfrac{{{{\left( {0.5} \right)}^2}}}{2} - \dfrac{{{0^2}}}{2}} \right]\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow F\left( {0.5} \right) = \dfrac{1}{{32}}\left[ {\dfrac{{0.25}}{2} - \dfrac{0}{2}} \right]\\ \Rightarrow F\left( {0.5} \right) = \dfrac{1}{{32}}\left[ {\dfrac{{\dfrac{1}{4}}}{2} - 0} \right]\\ \Rightarrow F\left( {0.5} \right) = \dfrac{1}{{32}}\left[ {\dfrac{1}{8}} \right]\end{array}\]
Multiplying the terms of the expression, we get
\[ \Rightarrow F\left( {0.5} \right) = \dfrac{1}{{256}}\]
Therefore, we get the value of the c.d.f. at \[x = 0.5\] as \[\dfrac{1}{{256}}\].
Now, we need to find the value of the cumulative density function at \[x = 9\].
Substituting \[x = 9\] in the c.d.f. \[F\left( x \right) = \int\limits_{ - \infty }^x {f\left( x \right)} dx\], we get
\[ \Rightarrow F\left( 9 \right) = \int\limits_0^9 {f\left( x \right)} dx\]
We can observe that 9 does not lie between 0 and 8.
We will split the limits using the property of definite integrals.
The definite integral \[\int\limits_a^b {f\left( x \right)} dx\] can be written as the sum of the definite integrals \[\int\limits_a^c {f\left( x \right)} dx\] and \[\int\limits_c^b {f\left( x \right)} dx\], where \[a < c < b\].
Therefore, we can rewrite the equation as
\[ \Rightarrow F\left( 9 \right) = \int\limits_0^8 {f\left( x \right)} dx + \int\limits_8^9 {f\left( x \right)} dx\]
The function in the first integral lies between 0 and 8. Therefore, \[f\left( x \right) = \dfrac{x}{{32}}\].
The function in the second integral does not lie between 0 and 8. Therefore, \[f\left( x \right) = 0\].
Therefore, the equation becomes
\[ \Rightarrow F\left( 9 \right) = \int\limits_0^8 {\dfrac{x}{{32}}} dx + \int\limits_8^9 {\left( 0 \right)} dx\]
Rewriting the equation, we get
\[ \Rightarrow F\left( 9 \right) = \int\limits_0^8 {\dfrac{x}{{32}}} dx + \int\limits_8^9 {\left( {0x} \right)} dx\]
The integral of a function of the form \[af\left( x \right)\] can be written as \[\int {af\left( x \right)} dx = a\int {f\left( x \right)} dx\].
Therefore, we get
\[\begin{array}{l} \Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\int\limits_0^8 x dx + 0 \times \int\limits_8^9 x dx\\ \Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\int\limits_0^8 x dx + 0\\ \Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\int\limits_0^8 x dx\end{array}\]
Rewriting the expression, we get
\[ \Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\int\limits_0^8 {{x^1}} dx\]
Therefore, we get
\[ \Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\left. {\left( {\dfrac{{{x^{1 + 1}}}}{{1 + 1}}} \right)} \right|_0^8\]
Simplifying the expression, we get
\[ \Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\left. {\left( {\dfrac{{{x^2}}}{2}} \right)} \right|_0^8\]
Substituting the limits into the expression, we get
\[ \Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\left[ {\dfrac{{{8^2}}}{2} - \dfrac{{{0^2}}}{2}} \right]\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\left[ {\dfrac{{64}}{2} - \dfrac{0}{2}} \right]\\ \Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\left[ {32 - 0} \right]\\ \Rightarrow F\left( 9 \right) = \dfrac{1}{{32}}\left[ {32} \right]\end{array}\]
Multiplying the terms of the expression, we get
\[ \Rightarrow F\left( 9 \right) = 1\]
Therefore, we get the value of the c.d.f. at \[x = 9\] as 1.

Note: Here we found out the cumulative density function. The cumulative density function gives the probability that a continuous random variable \[X\] takes a value less than or equal to \[x\]. A common mistake is to answer that the c.d.f. is 0 at \[x = 9\], and \[\dfrac{{0.5}}{{32}} = \dfrac{1}{{64}}\] at \[x = 0.5\]. This is incorrect because these are the values of the p.d.f. at the given values of \[x\].