Question

For a given discrete non-negative random variable, 𝑝𝑥 = 𝑃(𝑋 = 𝑥). Find 𝑥 such that 𝑝𝑥 > 𝑝𝑥+1, for the following distributions. a. 𝑋 is a binomial distribution with 𝑛 = 16 and 𝑝 = 0.5 b. 𝑋 is a Poisson distribution with 𝜆 = 8 c. 𝑋 is a Negative Binomial with 𝑟 = 5 and 𝑝 = 0.6 State the value of 𝑥 in the above sections. Verify whether the distributions are skewed for each section.

Answer 1

(a) 𝑋 is a binomial distribution with 𝑛 = 16 and 𝑝 = 0.5.

$p_x=P(X=x)=\frac{n!}{x!(n-x)!}p^xq^{n-x}$

$\frac{p_{x+1}}{p_x}=\left(\frac{n!}{(x+1)!(n-x-1)!}p^{x+1}q^{n-x-1}\right) /\left( \frac{n!}{x!(n-x)!}p^xq^{n-x}\right)=\frac{n-x}{x+1}\frac{p}{q}\leq 1$

$p(n-x)\leq q(x+1)$

$pn-q\leq x(p+q)=x$

$pn-q=0.5\cdot16-0.5=7.5$

The minimal integer $x\geq 7.5$ is 8.

Since $P(X=n-x)=\frac{n!}{x!(n-x)!}0.5^x0.5^{n-x}=\frac{n!2^{-n}}{x!(n-x)!}=P(X=x)$, any binomial distribution with $p=0.5$ is symmetric, hence, it is not skewed.

Answer. x=8

(b) 𝑋 is a Poisson distribution with 𝜆 = 8.

$p_x=P(X=x)=\frac{\lambda^{x}}{x!}e^{-\lambda}$

$\frac{p_{x+1}}{p_x}=\left(\frac{\lambda^{x+1}}{(x+1)!}e^{-\lambda}\right) /\left(\frac{\lambda^{x}}{x!}e^{-\lambda}\right)=\frac{\lambda}{x+1}\leq 1$

$x\geq \lambda-1=8-1=7$

The minimal integer $x\geq 7$ is 7, this means that the mode of X is equal to 7.

$E(X)=\sum\limits_{x=0}^{\infty}xP(X=x)=\sum\limits_{x=0}^{\infty}x\frac{\lambda^{x}}{x!}e^{-\lambda}=$

$=\lambda e^{-\lambda}\sum\limits_{x=1}^{\infty}\frac{\lambda^{x-1}}{(x-1)!}=\lambda e^{-\lambda}e^{\lambda}=\lambda=8$

Since $mode(X)=7<E(X)=8$, the probability distribution is skewed to the left.

Answer. x=8, the probability distribution is skewed to the left.

(c) 𝑋 is a Negative Binomial with 𝑟 = 5 and 𝑝 = 0.6

$p_x=P(X=x)=\frac{(x+r-1)!}{x!(r-1)!}q^x p^r$

$\frac{p_{x+1}}{p_x}=\left(\frac{(x+r)!}{(x+1)!(r-1)!}q^{x+1} p^r\right) /\left(\frac{(x+r-1)!}{x!(r-1)!}q^x p^r\right)=\frac{x+r}{x+1}q\leq1$

$qx+qr\leq x+1$

$qr-1\leq px$

$x\geq (qr-1)/p=(0.4\cdot 5-1)/0.6=5/3$

The minimal integer $x\geq 5/3$ is 2.

$E(X)=\sum\limits_{x=0}^{\infty}xP(X=x)=\sum\limits_{x=0}^{\infty}x\frac{(x+r-1)!}{x!(r-1)!}q^x p^r=$

$=\frac{q}{p}r\sum\limits_{x=1}^{\infty}\frac{((x-1)+(r+1)-1)!}{(x-1)!((r+1)-1)!}q^{x-1} p^{r+1}=\frac{q}{p}r=\frac{0.4}{0.6}5=10/3$

Since $mode(X)=2<E(X)=10/3$, the probability distribution is skewed to the left.

Answer. x=2, the probability distribution is skewed to the left.