91Ó°ÊÓ

Probability is the likelihood that an event will occur and is calculated by dividing the number of favourable outcomes by the total number of possible outcomes. The simplest example is a coin flip. When you flip a coin there are only two possible outcomes, the result is either heads or tails.

(a)

Let's break down the inequality:

\({X^2} \le y\)

It is required to put modulus when taking the root of an unknown quantity. Because \(y\) must be a positive number for the above inequality to hold at least once. As a result, after taking root on both sides,

We get,

\(|X| \le \sqrt y \)

Hence

\(\left\{ {{X^2} \le \sqrt y } \right\} = \{ |X| \le \sqrt y \} = \{ - \sqrt y \le X \le \sqrt y \} \)

Therefore, the solution of event is \(\left\{ {{X^2} \le \sqrt y } \right\} = \{ |X| \le \sqrt y \} = \{ - \sqrt y \le X \le \sqrt y \} \).

(b)

Because \(X\) has a regular normal distribution, its pdf is:

\(f(x) = \frac{1}{{\sqrt {2\pi } }} \cdot {e^{ - {x^2}/2}}\)

Hence the probability \(P\left( {{X^2} \le y} \right)\)can be written as :

\(P\left( {{X^2} \le y} \right) = P( - \sqrt y \le X \le \sqrt y )\)

\( = \int_{ - \sqrt y }^{\sqrt y } f (x) \cdot dx\)

\(P\left( {{X^2} \le y} \right) = \int_{ - \sqrt y }^{\sqrt y } {\frac{1}{{\sqrt {2\pi } }}} \cdot {e^{ - {x^2}/2}} \cdot dx\)

Now we'll use the following identity that has been given to us:

\(\frac{d}{{dy}}\left( {\int_{a(y)}^{b(y)} f (x) \cdot dx} \right) = f(b(y)) \cdot {b^\prime }(y) - f(a(y)) \cdot {a^\prime }(y)\)

Using this identity in the integral above we get :

\(P\left( {{X^2} \le y} \right) = \int_{ - \sqrt y }^{\sqrt y } {\frac{1}{{\sqrt {2\pi } }}} \cdot \exp - {x^2}/2 \cdot dx\)

\( = \left( {\frac{1}{{\sqrt {2\pi } }} \cdot {e^{ - {{(\sqrt y )}^2}/2}} \cdot \left( {\frac{1}{{2\sqrt y }}} \right)} \right) - \left( {\frac{1}{{\sqrt {2\pi } }} \cdot {e^{ - {{(\sqrt y )}^2}/2}} \cdot \left( {\frac{{ - 1}}{{2\sqrt y }}} \right)} \right)\)

\(P\left( {{X^2} \le y} \right) = \frac{1}{{\sqrt {2\pi } }} \cdot {y^{ - 1/2}} \cdot {e^{ - y/2}}\)

\({X^2}\) has a chi-squared distribution with \(\nu = 1\)df, as we can see from the cdf.