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. Find the mean and variance of the \(\beta\) distribution. Hint: From the pdf, we know that $$ \int_{0}^{1} y^{\alpha-1}(1-y)^{\beta-1} d y=\frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha+\beta)} $$ for all \(\alpha>0, \beta>0\)

Short Answer

Expert verified
The mean of the beta distribution is \(\mu = \frac{\alpha}{\alpha + \beta}\) and the variance is \(\sigma^2 = \frac{\alpha*\beta}{(\alpha + \beta)^2*(\alpha + \beta + 1)}\)

Step by step solution

01

Find the Mean

The mean \(\mu\) of the beta distribution is given by:\[\mu = \frac{\alpha}{\alpha + \beta}\]We can derive this by considering the expected value E[x] and applying the integral given in the problem statement, which says \[\int_{0}^{1} y^{\alpha-1}(1-y)^{\beta-1} d y=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}\]So, E[x] is basically \(\int_{0}^{1} x*y^{\alpha-1}(1-y)^{\beta-1} d y\). Now \(x\) can be fitted into the integral as \(x* y^{\alpha-1}\) or \( y^{\alpha}\). Substituting, we get \(\int_{0}^{1} (y^{\alpha}(1-y)^{\beta-1}) d y\) which is equal to \(\frac{\Gamma(\alpha + 1)\Gamma(\beta)}{\Gamma(\alpha+\beta + 1)}\). Using the properties of the Gamma function that says \(\Gamma(n + 1) = n*\Gamma(n)\), we get E[x] or \(\mu\) as \(\frac{\alpha}{\alpha + \beta}\) which is our mean.
02

Find the Variance

The variance \(\sigma^2\) is given by:\[\sigma^2 = \frac{\alpha*\beta}{(\alpha + \beta)^2*(\alpha + \beta + 1)}\]Here, we need to find E[x^2] similar to the way we found the mean. So, E[x^2] is basically \(\int_{0}^{1} x^2*y^{\alpha-1}(1-y)^{\beta-1} d y\) which becomes \(\int_{0}^{1} (y^{\alpha +1}(1-y)^{\beta-1}) d y\). Substituting from the hint again, we get \(\frac{\Gamma(\alpha + 2)\Gamma(\beta)}{\Gamma(\alpha+\beta + 2)}\). Now, using properties of Gamma function, we get E[x^2] as \(\frac{\alpha(\alpha +1)}{(\alpha + \beta)(\alpha + \beta +1)} = \frac{\alpha(\alpha + 1)}{(\alpha + \beta)^2(\alpha + \beta +1)}\).Now, the variance can be found using the formula \(\sigma^2 = E[x^2] - (E[x])^2\) which gives us the variance as \(\sigma^2 = \frac{\alpha*\beta}{(\alpha + \beta)^2*(\alpha + \beta + 1)}\) which is our variance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean of Beta Distribution
The mean of a Beta distribution is an essential measure of its central tendency. For a Beta distribution, the mean is succinctly expressed as \(\mu = \frac{\alpha}{\alpha + \beta}\).
Understanding the mean involves peering into the distribution's shape, which is defined by two parameters, \(\alpha\) and \(\beta\). These parameters dictate the skewness and weight of the distribution within the interval [0,1]. The ratio \(\frac{\alpha}{\alpha + \beta}\) encapsulates a balance point of the distribution, providing a valuable snapshot of its central location. To intuitively grasp this, imagine the curve of the Beta distribution on a graph. The mean signifies the point where the curve would balance if it were a physical object, with \(\alpha\) and \(\beta\) contributing to the tilt towards 0 or 1. This balance point describes the 'expected' outcome, a term frequently used in probability and statistics to represent the mean.
Variance of Beta Distribution
Variance quantifies the spread of the Beta distribution around its mean. It's represented by\(\sigma^2 = \frac{\alpha*\beta}{(\alpha + \beta)^2*(\alpha + \beta + 1)}\).
To understand variance, one must appreciate its role in highlighting the dispersion of a distribution. Variance is always non-negative, and a larger value denotes a wider, more dispersed distribution. The Beta distribution is unique in that its variance is bounded, thanks to the distribution's support being confined to the interval [0,1]. The formula encapsulates the influence of both \(\alpha\) and \(\beta\), and as these parameters increase, the distribution becomes more 'peaked', reducing the overall variance. It conveys how tightly data points are clustered around the mean, which is imperative when evaluating risk and uncertainty in various applications, such as finance or quality control.
Gamma Function
The Gamma function is a cornerstone in understanding the Beta distribution and provides a bridge to computing various statistical measures. It generalizes factorial to non-integer values, satisfying\(\Gamma(n + 1) = n*\Gamma(n)\)for positive n. In the realm of the Beta distribution, the Gamma function is pivotal because it appears in the normalization constant ensuring the total probability integrates to one. Its properties allow us to transition from factorials to continuous parameters \(\alpha\) and \(\beta\) in computational forms, crucial when calculating moments like the mean and variance of the Beta distribution. Think of the Gamma function as a versatile tool, similar to factorials, but with a broader scope for continuous calculus operations, providing the necessary flexibility for complex probability distributions.
Expected Value in Probability
The expected value is the backbone of probability theory, effectively conveying an average outcome of a probabilistic scenario. It's defined mathematically as\(E[x] = \int_{-\infty}^{\infty} x*f(x) dx\),for a continuous variable, where \(f(x)\) is the probability density function. For the Beta distribution, we compute the expected value by integrating over the product of the value and its probability, which leads to the calculation of the mean. This concept is pivotal as it provides a concrete measure to predict the 'center' of mass of a distribution, where outcomes are likely to concentrate around given repeated sampling. The expected value guides decision-making in various fields, such as economics and insurance, by quantifying potential risks and rewards, fundamentally shaping strategies and anticipations.

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Most popular questions from this chapter

Let the number of chocolate drops in a certain type of cookie have a Poisson distribution. We want the probability that a cookie of this type contains at least two chocolate drops to be greater than \(0.99 .\) Find the smallest value of the mean that the distribution can take.

Show that the moment generating function of the negative binomial distribution is \(M(t)=p^{r}\left[1-(1-p) e^{t}\right]^{-r}\). Find the mean and the variance of this distribution. Hint: In the summation representing \(M(t)\), make use of the MacLaurin's series for \((1-w)^{-r}\)

Let \(F\) have an \(F\) -distribution with parameters \(r_{1}\) and \(r_{2} .\) Prove that \(1 / F\) has an \(F\) -distribution with parameters \(r_{2}\) and \(r_{1}\).

. Let \(X\) have a Poisson distribution with parameter \(m .\) If \(m\) is an experimental value of a random variable having a gamma distribution with \(\alpha=2\) and \(\beta=1\), compute \(P(X=0,1,2)\). Hint: Find an expression that represents the joint distribution of \(X\) and \(m\). Then integrate out \(m\) to find the marginal distribution of \(X\).

. In the Poisson postulates of Remark 3.2.1, let \(\lambda\) be a nonnegative function of \(w\), say \(\lambda(w)\), such that \(D_{w}[g(0, w)]=-\lambda(w) g(0, w) .\) Suppose that \(\lambda(w)=\) \(k r w^{r-1}, r \geq 1\) (a) Find \(g(0, w)\) noting that \(g(0,0)=1\). (b) Let \(W\) be the time that is needed to obtain exactly one change. Find the distribution function of \(W\), i.e., \(G(w)=P(W \leq w)=1-P(W>w)=\) \(1-g(0, w), 0 \leq w\), and then find the pdf of \(W\). This pdf is that of the Weibull distribution, which is used in the study of breaking strengths of materials.

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