Problem 61

Question

A continuous random variable $Y$ has a cdf given by $$ F_{Y}(y)= \begin{cases}0 & y<0 \\ y^{2} & 0 \leq y<1 \\ 1 & y \geq 1\end{cases} $$ Find \(P\left(\frac{1}{2}

Step-by-Step Solution

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Answer

The probability of Y being in the range $ \frac{1}{2} < y \leq \frac{3}{4}$ is 9/32 through both methods of using the cdf and pdf.

1Step 1: Find Probability Using CDF

To find the probability using the cdf, direct substitution and evaluation is needed. The function $F_{Y}(y)=y^{2}$ is used since $ \frac{1}{2} < y \leq \frac{3}{4}$ lies in the range where this function is defined. To do this, calculate the difference $ F_{Y}\left(\frac{3}{4}\right) - F_{Y}\left(\frac{1}{2}\right)$ which calculates the cumulative probability up to 3/4 and subtracts the cumulative probability up to 1/2. This effectively gives the probability of the variable Y being in the range $ \frac{1}{2} < y \leq \frac{3}{4}$.

2Step 2: Find Probability Using PDF

First, the pdf needs to be obtained from the cdf. The pdf is the derivative of the cdf. Thus, obtain the pdf by differentiating the cdf function $ F_Y(y) = y^{2}$ to get $ f_Y(y) = 2y $ for $ 0 \leq y < 1 $. Calculate the integral of the pdf from 1/2 to 3/4. The operation of the integral sums up the infinitesimal probabilities of the random variable Y being in the range of 1/2 to 3/4, resulting in the total probability of Y being in that range.

Key Concepts

Cumulative Distribution Function (cdf)Probability Density Function (pdf)Probability Calculation

Cumulative Distribution Function (cdf)

The Cumulative Distribution Function (cdf) is an essential concept when dealing with continuous random variables. It's a function that describes the probability that a random variable will take a value less than or equal to a certain point. In simpler terms, the cdf gives us the cumulative probability up to a specific value.For example, consider the given continuous random variable, denoted as $Y$, with the cdf defined piecewise:

$F_{Y}(y)=0$ for $y<0$
$F_{Y}(y)=y^2$ for $0 \leq y<1$
$F_{Y}(y)=1$ for $y \geq 1$

This means that for values of $y$ between 0 and 1, the cdf is a quadratic function $y^2$. We calculate probabilities by finding the difference in cdf values at the endpoints of the desired range. For example, to find \(P\left(\frac{1}{2}

Probability Density Function (pdf)

A Probability Density Function (pdf) is derived from the cumulative distribution function (cdf) for a continuous random variable. It represents the probability of the random variable falling within a particular range of values, rather than at one specific value.The pdf is essentially the derivative of the cdf. Derivatives represent the rate of change of a function, so differentiating the cdf provides the likelihood of the variable landing exactly at a particular small interval. For our given random variable $Y$, since the cdf for $0 \leq y < 1$ is $y^2$, the pdf $f_Y(y)$ is the derivative of $y^2$. Thus, $f_Y(y) = 2y$.The pdf allows us to calculate probabilities over a range by integrating over that range. For $Y$ in the range from 1/2 to 3/4, we integrate the pdf:\[\int_{1/2}^{3/4} 2y \, dy\]This integral sums up the "density" of probabilities across this interval, providing the probability of the event occurring within this range.

Probability Calculation

Probability calculation using cdf and pdf provide two methods to find the probability of a random variable being within a specific range.

Using the cdf

To find probabilities using the cdf, we calculate the difference between the cumulative probabilities at the boundaries of the interval. For example, to find \(P\left(\frac{1}{2}

Find $F_{Y}\left(\frac{3}{4}\right) = \left(\frac{3}{4}\right)^2 = \frac{9}{16}$

Find $F_{Y}\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$

Subtract: $\frac{9}{16} - \frac{1}{4} = \frac{9}{16} - \frac{4}{16} = \frac{5}{16}$

Using the pdf

Alternatively, use the pdf to find the probability by integrating over the interval. Consider the pdf $f_Y(y) = 2y$:

Integrate: \[\int_{1/2}^{3/4} 2y \, dy = \left[y^2\right]_{1/2}^{3/4} = \left(\frac{9}{16} - \frac{1}{4}\right) = \frac{5}{16}\]

Both methods converge to the same result, ensuring consistency in probability calculations.

Problem 60

Problem 62

Other exercises in this chapter

Problem 59

If $Y$ is an exponential random variable, $f_{Y}(y)=$ $\lambda e^{-\lambda y}, y \geq 0$, find $F_{Y}(y)$

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Problem 60

If the pdf for $Y$ is $$ f_{Y}(y)= \begin{cases}0, & |y|>1 \\ 1-|y|, & |y| \leq 1\end{cases} $$ find and graph $F_{Y}(y)$.

View solution

Problem 62

A random variable $Y$ has cdf $$ F_{Y}(y)= \begin{cases}0 & y

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Problem 63

The cdf for a random variable $Y$ is defined by $F_{Y}(y)=0$ for $y1$. Find $P\left(\frac{1}{4} \leq Y \leq \frac{3}{4}\right)$ by integrating \(f_{Y}(y

View solution