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Please help and provide suggestions.

(i) A discrete random variable X has the distribution $P(X = i) = 2a^i $ for i ∈ N+ (where N+ := {1,2,...}). What is the value of a?

$P(X = 1) = 2a^1 ;P(X = 2) = 2a^2 ;P(X = 1) = 2a^3 ;P(X = 1) = 2a^4;...... $

$S_n=\sum_{i=1}^\infty 2a^i=\frac{2a(1-a^n)}{1-a}=1\iff2a-2a^{n+1}=1-a\iff 2a^{n+1}-3a+1=0\iff a=1$

Is this correct?

(ii) A random variable X is said to follow the Cauchy distribution if it’s density function $f_X(x)$ is given by $$f_X(x)= \frac1π\frac{1}{(1+x^2)}$$Show that $f_X (x)$ is a valid probability density function and compute the variance of X.

$\int_{-\infty} ^\infty \frac1π\frac{1}{(1+x^2)}=\frac1π[tan^{-1}x]_{-\infty}^\infty=\frac1π[\frac\pi2-\frac{-\pi}{2}]=1$

$Var(X)=E(X^2)-E(X)^2$

I've got no idea but is this right.

$E(X)=\int_{-\infty} ^\infty \frac1π\frac{x}{(1+x^2)}=\frac{1}{2\pi}ln(\frac{1+(\infty)^2}{1+(-\infty)^2})=\frac{1}{2\pi}ln(\frac\infty \infty) $ is undefined

$E(X^2)=\int_{-\infty} ^\infty \frac1π\frac{x^2}{(1+x^2)}=\frac1π\int_{-\infty} ^\infty 1+\frac{-1}{(1+x^2)}=\frac{1}{\pi}[x-tan^{-1}x]_{-\infty}^{\infty}=\frac{1}{\pi}[(\infty-\frac{\pi}{2})-(-\infty-\frac{-\pi}{2})]=\frac{1}{\pi}[2\infty-1]=\infty$

Therfore, $Var(X)=E(X^2)-E(X)^2=\infty -(undefined)^2=undefined$

user117498
  • 775

2 Answers2

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As $0\le 2a\le 1$ then $a^n\to 0$ as $n\to \infty$. So $$1=S=\sum_{i=1}^\infty 2a^i=\frac{2a}{1-a}.$$ So $2a=1-a$ and then $a=\frac{1}{3}$.

Mohammad Khosravi
  • 2,333
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Part 1 is incorrect..Think about it. If you let a=1, then the summation of all the individual probabilities will be greater than 1. It should be more like $2(a)/(1-a) -2=1$ Therefore $\frac{3}{2}-\frac{3}{2}a=a$ and $a=\frac{3}{5}$

Tom
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