Let $f (x)$ be the pdf of $X,$ hence from deﬁnition of expected value of a random variable we write

∫ ∞ E (X ) = xf (x )dx − ∞

Now break the integral into the sum of integrals as follows

∫ ξ−δ ∫ ξ ∫ ξ+ δ ∫ ξ+2δ E (X ) = ⋅⋅ ⋅ + xf (x )dx + xf (x )dx + xf (x)dx + xf (x)dx + ⋅⋅⋅ ξ−2δ ξ−δ ξ ξ+ δ

In the limit, as $δ$ is made very small, the above can be written as Riemann sums of areas each of width $dx → δ$ as follows

But due to symmetry around $ξ$ then

f (ξ − iδ) = f (ξ + iδ)

for any integer $i$ in the above Riemann sum. This causes terms to cancel in the equation (1) above.

For example the term $− δf (ξ − δ)$ onthe left of the $ξf ξ$ will cancel with the term $+ δf (ξ − δ)$ on the right of $ξfξ$ , and so on. Then we obtain the following sum

E (X ) = δ[⋅⋅⋅ + ξf (ξ − 2δ) + ξf (ξ − δ) + ξfξ + ξf (ξ + δ) + ξf (ξ + 2δ) + ⋅⋅⋅]

Take $ξ$ as common factor

E (X ) = ξδ [⋅⋅⋅ + ξf (ξ − 2δ) + ξf (ξ − δ) + ξfξ + ξf (ξ + δ) + ξf (ξ + 2δ) + ⋅⋅⋅]

(2)

But

δ[⋅⋅⋅ + ξf (ξ − 2δ) + ξf (ξ − δ) + ξfξ + ξf (ξ + δ ) + ξf (ξ + 2δ) + ⋅⋅⋅]

is just the total area under $f (x)$ in the Riemann sum sense i.e. $∫ ∞ − ∞ f (x )dx$ .

Hence (2) becomes

∫ ∞ E (X ) = ξ f (x)dx −∞

But since $f (x)$ is a density, this area is one. Hence

|----------| E-(X-)-=-ξ--

The density function of an exponential distribution with parameter $λ$ is given by

{ −λx f (x) = λe x ≥ 0 0 x < 0

First ﬁnd the expected values of an exponential random variable $X.$ From deﬁnition of expected value:

integrate by parts gives

Hence $1 E (X ) = λ$ , Hence we need to ﬁnd $(|| 1|| 2) Δ = P X − λ > λ$ , But this is the same as ﬁnding

Now compare to Chebyshev bound. Chebyshev bound says that

Var (X ) P (|X − E (X )| ≥ t) ≤ ----2---- t

(1)

Hence the upper bound by Chebyshev is $V-ar(X) (2λ)2$ . We now need to ﬁnd $V ar(X )$ and this is given by

( ) Var (X ) = E X2 − [E (X )]2

But

Hence (1) becomes

( ) 1- |----| P |X − E (X )| ≥ 2- ≤ λ2-= -0.25- λ 4λ2

Hence an upper bound for the probability by Chebyshev is $0.25$ , and the actual probability found was $0.049787$ which is well within this bound.

Let $∑∞ Δ = P (X ≥ K ) k=1$ , we need to show that this equals $E (X )$

But

P (X ≥ 1) = P (X = 1) + P (X = 2) + P (X = 3) + ⋅⋅⋅

and

P (X ≥ 2) = P (X = 2) + P (X = 3) + P (X = 4) + ⋅⋅⋅

and

P (X ≥ 3) = P (X = 3) + P (X = 4) + P (X = 5) + ⋅⋅⋅

and so on. Hence adding all the above we obtain repeated terms, which comes out as follows

But this is the deﬁnition of $E (X )$ , hence $Δ = E (X )$

$X$ is Number of trials needed to obtain $r$ successes, Each trial has $p$ chance of success.

Let $Y1$ be a random variable which represents the number of trials to obtain a success (counting the success trial) (This will be the ﬁrst success).

Let $Y 2$ be a random variable which represents the number of trials to obtain a success (this will be the second success so far)

Let $Y3$ be a random variable which represents the number of trials to obtain a success (this will be the third success so far)

and so on. Hence

Let $Y i$ be a random variable which represents the number of trials to obtain the $ith$ success.

Therefore

Hence

But a Geometric r.v. represents the number of trials needed to obtain a success (counting the success trial), with each trial having p chance of success. So we need to ﬁnd $E (Y )$ where $Y$ is a Geometric r.v. with parameters $p$