Univariate distributions

Activity 3.1   Show that for a discrete random variable $X$ taking integer values $x$, $P(X=x)=F_X(x)-F_X(x-1)$.
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Answer 3.1   The event ${X\le x}={X=x}\cup{X \le x-1}$, and the two events on the right are disjoint, so using the addition rule,

$$P(X\le x)=P(X=x)+P(X\le x-1),$$

and this may be rewritten

$$F_X(x)=P(X=x)+F_X(x-1),$$

which immediately gives the result wanted.
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Activity 3.2   The greengrocer has a very large (effectively infinite) pile of oranges on his stall. The pile of fruit is a mixture of 50% old fruit with 50% new fruit; one can't tell which are old and which are new. However, 20% of old oranges are mouldy inside, but only 10% of new oranges are mouldy. Suppose that you choose 5 oranges at random. Is it true that the number of mouldy oranges in your sample has a binomial distribution with $n=5$ and $\pi=0.15$?
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Answer 3.2   For an orange chosen at random, the event ${\text{orange is mouldy}}$ is the union of the two disjoint events ${\text{orange is mouldy}\cap \text{orange is new}}$ and ${\text{orange is mouldy}\cap \text{orange is old}}$. So

$$P( ) =P( \cap )+P( \cap )$$

$$=P( \vert )P( )+P( \vert )P( )$$

$$=0.1\times 0.5 +0.2\times 0.5=0.15.$$

As the pile of oranges is very large, the results for the five oranges will be independent, so we have 5 independent trials each with probability of mouldy equal to 0.15. The distribution of the number of mouldy oranges will be a binomial distribution with $n=5$ and $\pi=0.15$.
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Activity 3.3   The chance that a lottery ticket has a winning number is .0000001. If 10,000,000 people buy tickets that are independently numbered, what is the probability of 0 winner? Of 1 winner? (Hint, use a Poisson distribution with mean 1. The answers are 0.37, 0.37.)
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Answer 3.3   We would expect in this case that the Poisson distribution would be a good approximation to the binomial distribution with $n=10^7$ and $\pi=1.0\times 10^{-7}$ that would be be the natural distribution to use. The approximation is good because $\pi$ is very small and $n$ very large. The mean for the approximating Poisson distribution is $\mu=n\pi=1$. From the Poisson Tables, Table 2 of the New Cambridge Statistical Tableswith $\mu = 1.00$ and $r=1$, $r=0$ we get

$$P(X\le 1) =0.7358.$$

$$P(X\le 0) = 0.3679.$$

So the probability of exactly 1 winner is $0.7358-0.3679 = 0.3679$, and of exactly 0 winner is the same as $P(X\le 0) = 0.3679$.

One could also use the probability function for the Poisson distribution.

$$P(X=0)=e^{-1} \frac{1^0}{0!}=e^{-1}\frac{1}{1}=e^{-1}=0.3679.$$

$$P(X=1)=e^{-1} \frac{1^1}{1!}=e^{-1}\frac{1}{1}=e^{-1}=0.3679.$$

If we use the binomial distribution we get

$$P(X=0)=\binom{10^7}{0}(10^{-7})^0(1-10^{-7})^{10^7}=(1-10^{-7})^{10^7}=0.3679,$$

$$P(X=0)=\binom{10^7}{1}(10^{-7})^1(1-10^{-7})^{10^7-1}=(1-10^{-7})^{10^7-1}=0.3679.$$

The Poisson approximation is correct to at least 4 decimal places. Notice that if one wanted, say, $P(X=5)$ it would become difficult to do the binomial calculation.
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Activity 3.4   Show that for a uniform random variable on $(0,1)$, if the probability $P(X=x)$ is the same for all $x$ between 0 and $1$, then it must be equal to 0 for all those $x$ (otherwise the probability of the sample space is not equal to $1$, but infinite).
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Answer 3.4   Suppose that $P(X=x)=c$ for all choices of $x$ between 0 and 1. We can define a disjoint events $A_i$, for $i=1,2,\dots$ by $A_i={1/2^i}$. From the second axiom of probability, for each positive integer $n$.

$$P(A_1\cup A_2\cup A_3 \cup ...\cup A_n) = P(A_1)+P(A_2) +P(A_3) + \dots +A_n$$

$$=c+c+\dots...+c=nc.$$

For each choice of $n$ the probability on the left must, from the first axiom of probability be less than or equal to 1, so that $nc$ must be less than or equal to 1 for each choice of $n$. Unless $c=0$ this is not true, because for each given $c >0$ we can always choose $n$ so large that $nc>1$. For instance, suppose we think that $c=10^{-6}$ might do. It will not, because by choosing $n=10^7$ we would get $nc=10$, and would have constructed an event with probability greater than 1.
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Activity 3.5   Suppose that $X$ is uniformly distributed on $(0,1)$.
What is $P(X>0.2)$, $P(X\ge 0.2)$, $P(X^2>0.04)$. (Hint: All these probabilities are the same.)
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Answer 3.5   We have $a=0,b=1$, and can use the formula for $P(c< X \le d)$.

$$P(X>0.2)=P(0.2<X\le 1)=\frac{1-0.2}{1-0}=0.8.$$

$$P(X\ge 0.2)= P(X=0.2)+P(X>0.2)=0+P(X>0.2).$$

$$P(X\ge 0.2)= P(X=0.2)+P(X>0.2)=0+P(X>0.2).$$

$$P(X^2> 0.4)= P(X<-0.2)+P(X>0.2)=0+P(X>0.2).$$

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Activity 3.6   If $X$ is a random variable with a standard normal distribution, what is $P(X^2>3.84)$?
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Answer 3.6  

$$PX^2>3.84) =P(X < -\sqrt{3.84})+P(X > \sqrt{3.84})$$

$$=P(X < -1.96)+P(X > 1.96)$$

$$=2(1-\Phi(1.96))=2(1-0.975)=0.05.$$

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Activity 3.7   Show that the mean of the exponential distribution is $1/\lambda$.
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Answer 3.7   If $X$ has density function $f_X(x) = \lambda e^{-\lambda x}$, then the mean of $X$ is

$$\int_0^\infty x\lambda e^{-\lambda x}dx=[-xe^{-\lambda x} - e^{-\lambda x}/\lambda]^\infty_0 = 1/\lambda.$$

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Activity 3.8   Find the mean of the Bernoulli trial distribution, where $X=0$ with probability $1-\pi$ and $X=1$ with probability $\pi$.
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Answer 3.8   The mean of $X$ is

$$E(X)=0\times (1-\pi)+1\times \pi=\pi.$$

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Activity 3.9   What is the expected value of $X$ if the only possible value of $X$ is 0? (We have $P(X=0)=1$, so $X$ is effectively a constant, even though it is called a random variable.)
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Answer 3.9  

$$EX=0\times P(X=0)=0\times 1=0.$$

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Activity 3.10   Show, that $E[X-E(X)]=0$.
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Answer 3.10   $X-E(X)$ is the sum of two random variables $X$ and the degenerate random variable $-E(X)$ that takes the value $-E(X)$ with probability 1. so,

$$E[X+(-E(X))]=E(X)+E[-E(X)]=E(X)+(-E(X))=0.$$

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Activity 3.11   Show that the variance of the Bernoulli trial distribution is $\pi(1-\pi)$.
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Answer 3.11   We know already that $E(X)=\pi$, and

$$E(X^2)=0^2\times P(X=0)+1^2\times P(X=1)=p(X=1)=\pi.$$

So $\operatorname{var}X=E(X^2)-[E(X)]^2=\pi-\pi^2=\pi(1-\pi)$.
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Activity 3.12   Show that if $\operatorname{var}X =0$ then $P(X=\mu_X)=1$. (We say in this case that $X$ is almost surely equal to its mean.)
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Activity 3.13   Find the variance of the exponential distribution with mean $1/\lambda$.
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Answer 3.12   We know already that $E(X)=1/\lambda$. $E(X^2)$ is

$$\int_0^\infty x\lambda e^{-\lambda x}dx =[-x^2e^{-\lambda x}]^\infty_0 +2\int_0^\infty xe^{-\lambda x}dx$$

$$= [-x^2e^{-\lambda x}-2xe^{-\lambda x}/\lambda ]^\infty_0+ 2\int_0^\infty e^{-\lambda x}/\lambda$$

$$=[-x^2e^{-\lambda x}-2e^{-\lambda x}/\lambda+-2e^{-\lambda x}/\lambda^2]^\infty_0$$

$$=2/\lambda^2$$

So the variance is $2/\lambda^2 -1/\lambda^2 =1/\lambda^2$.
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