# Deferred Norlund statistical convergence in probability, mean and
distribution for sequences of random variables

Kuldip Raj, Swati Jasrotia  School of Mathematics Shri Mata Vaishno Devi
University, Katra-182320, J & K (India<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Reza Saadati∗ Reza Saadati
School of Mathematics, Iran University of Science and Technology, Narmak,
Tehran, Iran<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We introduce and study deferred Nörlund statistical convergence in
probability, mean of order $r,$ distribution and study the interrelation among
them. Based upon the proposed method to illustrate the findings, we presented
new Korovkin type theorems for the sequence of random variables via deferred
Nörlund statistically convergence and present compelling examples to
demonstrate the effectiveness of the results.

###### Key words and phrases:

Probability convergence, deferred Nörlund, mean convergence, distribution
convergence, statistical convergence

###### 2010 Mathematics Subject Classification:

40A05, 40A30

##  1\. Introduction and Preliminaries

Fast [5] and also Schoenberg [7] studied the concept of statistical
convergence and continued by Rath–Tripathy [8] and Gadjiev–Orhan [20].

Suppose that $(x_{m})$ and $(y_{m})$ are the sequences of non-negative
integers fulfilling

(1.1)
$x_{m}<y_{m},\;\forall\;m\;\in\mathbb{N}\;\;\;\;\;\;\;\;\;\;\;\;\mbox{and}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\lim_{x\rightarrow\infty}y_{m}=\infty.\\\
$

Further, let $(e_{m})$ and $(g_{m})$ be two sequences of non-negative real
numbers such that

(1.2)
$\mathcal{E}_{m}=\displaystyle\sum_{n=x_{m}+1}^{y_{m}}e_{n}\;\;\;\mbox{and}\;\;\;\mathcal{F}_{m}=\displaystyle\sum_{n=x_{m}+1}^{y_{m}}g_{n}.\\\
$

The convolution of (1.2) is defined as

$\mathcal{R}_{m}=\displaystyle\sum_{v=x_{m}+1}^{y_{m}}e_{v}g_{y_{m}-v}.$

As introduced by Srivastava et al. in [24], the deferred Nörlund $(DN)$ mean
is defined as

$t_{m}=\frac{1}{\mathcal{R}_{m}}\displaystyle\sum_{n=x_{m}+1}^{y_{m}}e_{y_{m}-n}g_{n}y_{n}.$

Suppose that $(x_{m})$ and $(y_{m})$ are the sequences fulfilling conditions
(1.1) and $(e_{m}),(g_{m})$ are sequences satisfying (1.2). A sequence
$(Y_{m})$ is called as deferred Nörlund statistically convergent to $Y$ if
$\forall\;\varepsilon>0,$ the set

$\\{n:n\leq\mathcal{R}_{m},\;\mbox{and}\;e_{y_{m}-n}g_{n}|Y_{m}-Y|\geq\varepsilon\\}$

has zero deferred Nörlund density, i.e. if

$\lim_{m\rightarrow\infty}\frac{1}{\mathcal{R}_{m}}\Big{|}\Big{\\{}n:n\leq\mathcal{R}_{m}\;\;\mbox{and}\;\;e_{y_{m}-n}g_{n}|Y_{m}-Y|\geq\varepsilon\Big{\\}}\Big{|}=0.$

We write it as

$St_{DN}\lim Y_{m}=Y.$

Suppose that $(x_{m})$ and $(y_{m})$ are the sequences fulfilling conditions
(1.1) and $(e_{m}),(g_{m})$ are sequences satisfying (1.2). A sequence
$(Y_{m})$ is called as deferred Nörlund statistically probability (or
$St_{DNP}-$) convergent to a random variable $Y,$ if $\forall\;\varepsilon>0$
and $\delta>0,$ the set

$\\{n:n\leq\mathcal{R}_{m}\;\;\mbox{and}\;\;e_{y_{m}-n}g_{n}P(|Y_{m}-Y|\geq\varepsilon)\geq\delta\\}$

has $DN-$density zero, i.e.,

$\displaystyle\lim_{m\rightarrow\infty}\frac{1}{\mathcal{R}_{m}}\Big{|}\Big{\\{}n:n\leq\mathcal{R}_{m}\;\;\mbox{and}\;\;e_{y_{m}-n}g_{n}P(|Y_{m}-Y|\geq\varepsilon)\geq\delta\Big{\\}}\Big{|}=0$

or

$\displaystyle\lim_{m\rightarrow\infty}\frac{1}{\mathcal{R}_{m}}\Big{|}\Big{\\{}n:n\leq\mathcal{R}_{m}\;\;\mbox{and}\;\;1-e_{y_{m}-n}g_{n}P(|Y_{m}-Y|\leq\varepsilon)\geq\delta\Big{\\}}\Big{|}=0,$

and it is denoted as

$\displaystyle
St_{DNP}\lim_{m\rightarrow\infty}e_{y_{m}-n}g_{n}P(|Y_{m}-Y|\geq\varepsilon)$
$\displaystyle=$ $\displaystyle 0$

or

$\displaystyle
St_{DNP}\lim_{m\rightarrow\infty}e_{y_{m}-n}g_{n}P(|Y_{m}-Y|\leq\varepsilon)$
$\displaystyle=$ $\displaystyle 1.$

##  2\. Deferred Nörlund statistically probability convergence

In this section, we study deferred Nörlund statistically probability
convergence, for a historical review and basic concept we refer [3], [4],
[10], [12], [19] [9], [15], [23], [16], [17] and [27].

######  Theorem 2.1.

Suppose that $(Y_{m})$ and $(Z_{m})$ are sequences of random variables and
consider two random variables $Y$ and $Z.$ Then the following assertions are
satisfied

  1. (1)

$St_{DNP}Y_{m}\rightarrow Y$ and $St_{DNP}Y_{m}\rightarrow Z\Rightarrow
P(Y=Z)=1,$

  2. (2)

$St_{DNP}Y_{m}\rightarrow y\Rightarrow St_{DNP}Y^{2}_{m}=y^{2},$

  3. (3)

$St_{DNP}Y_{m}\rightarrow y$ and $St_{DNP}Z_{m}\rightarrow z\Rightarrow
St_{DNP}Y_{m}Z_{m}\rightarrow yz,$

  4. (4)

$St_{DNP}Y_{m}\rightarrow y$ and $St_{DNP}Z_{m}\rightarrow z\Rightarrow
St_{DNP}\frac{Y_{m}}{Z_{m}}\rightarrow\frac{y}{z},\;z\neq 0,$

  5. (5)

$St_{DNP}Y_{m}\rightarrow Y$ and $St_{DNP}Z_{m}\rightarrow Z\Rightarrow
St_{DNP}Y_{m}Z_{m}\rightarrow YZ,$

  6. (6)

if $St_{DNP}Y_{m}\rightarrow Y\;\forall\;\varepsilon,\delta>0,$ then
$\exists\;a\in\mathbb{N}$ s.t.

$d(\\{n:n\leq\mathcal{R}_{m}\;\;\mbox{and}\;\;e_{y_{m}-n}g_{n}P(|Y_{m}-Y_{a}|\geq\varepsilon)\\\
\geq\delta\\})=0.$

###### Proof.

Let $\epsilon$ and $\delta$ be positively small real numbers. Also consider
$(x_{m})$ and $(y_{m})$ are the sequences fulfilling conditions (1.1) and
$(e_{m}),(g_{m})$ are sequences satisfying (1.2).

  1. (1)

Suppose that
$a\in\Big{\\{}n:n\leq\mathcal{R}_{m}\;\;\mbox{and}\;\;e_{y_{m}-n}g_{n}P\Big{(}|Y_{m}-Y|\geq\frac{\varepsilon}{2}\Big{)}<\frac{\delta}{2}\Big{\\}}\cap\Big{\\{}n:n\leq\mathcal{R}_{m}\;\;\mbox{and}\;\;e_{y_{m}-n}g_{n}P\Big{(}|Y_{m}-Z|\geq\frac{\varepsilon}{2}\Big{)}<\frac{\delta}{2}\Big{\\}}$
(as the limit density of both the sets is $1$). Then,
$e_{y_{m}-n}g_{n}P\Big{(}|Y-Z|\geq\varepsilon\Big{)}\leq
e_{y_{m}-n}g_{n}P\Big{(}|Y_{a}-Y|\geq\frac{\varepsilon}{2}\Big{)}+e_{y_{m}-n}g_{n}P\Big{(}|Y_{a}-Z|\geq\frac{\varepsilon}{2}\Big{)}<\delta.$
It means

$P\\{Y=Z\\}=1.$

  2. (2)

If $St_{DNP}Y_{m}\rightarrow 0$, then $St_{DNP}Y^{2}_{m}\rightarrow 0.$ Here,
we see that
$a\in\\{n:n\leq\mathcal{R}_{m}\;\;\mbox{and}\;\;e_{y_{m}-n}g_{n}P(|Y_{m}-0|\geq\varepsilon)>\delta\\}=a\in\\{n:n\leq\mathcal{R}_{m}\;\;\mbox{and}\;\;e_{y_{m}-n}g_{n}P(|Y^{2}_{m}-0|\geq\varepsilon>\delta\\}.$
Now, take $Y_{m}^{2}=(Y_{m}-y)^{2}+2y(Y_{m}-y)+y^{2}.$ Thus,
$St_{DNP}Y_{m}^{2}\rightarrow y^{2}.$

  3. (3)

Suppose that $St_{DNP}Y_{m}\rightarrow y$ and $St_{DNP}Z_{m}\rightarrow z.$ As
$St_{DNP}Y_{m}Z_{m}=St_{DNP}\frac{1}{4}\\{(Y_{m}+Z_{m})^{2}-(Y_{m}-Z_{m})^{2}\\}=\frac{1}{4}\\{(y_{m}+z_{m})^{2}-(y_{m}-z_{m})^{2}\\}=yz.$

  4. (4)

Suppose that $R$ and $S$ be two events correspond $|Z_{m}-z|<|z|$ and
$\Big{|}\frac{1}{Z_{m}}-\frac{1}{z}\Big{|}\geq\varepsilon.$ We have

$\Big{|}\frac{1}{Z_{m}}-\frac{1}{z}\Big{|}=\frac{|Z_{m}-z|}{|zZ_{m}|}=\frac{|Z_{m}-z|}{|z|\cdot|z+(Z_{m}-z)|}\leq\frac{|Z_{m}-z|}{|z|\cdot|(|z|-|Z_{m}-z|)|}.$

If the events $R$ and $S$ occurs at same time, then

$|Z_{m}-z|\geq\frac{\varepsilon|z^{2}|}{1+\varepsilon|z|}.$

Further, let $\varepsilon_{0}=\varepsilon|z|^{2}/(1+\varepsilon|z|)$ and $A$
be the event such that $|Z_{m}-z|\geq\varepsilon_{0}.$ Thus,

$RS\subseteq A\Rightarrow P(S)\leq P(A)+P(R^{c}).$

Thus,
$\Big{\\{}n:n\leq\mathcal{R}_{m}\;\;\mbox{and}\;\;e_{y_{m}-n}g_{n}P\Big{(}|\frac{1}{Z_{m}}-\frac{1}{z}|\geq\varepsilon\Big{)}\geq\delta\Big{\\}}\subseteq\Big{\\{}n:n\leq\mathcal{R}_{m}\;\;\mbox{and}\;\;$

$\displaystyle\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;e_{y_{m}-n}g_{n}P\big{(}|Z_{m}-z|\geq\varepsilon_{0}\big{)}\geq\frac{\delta}{2}\Big{\\}}\cup\Big{\\{}n:n\leq\mathcal{R}_{m}\;\;\mbox{and}\;\;e_{y_{m}-n}g_{n}$
$\displaystyle P\big{(}|Z_{m}-z|\geq|z|\big{)}\geq\frac{\delta}{2}\Big{\\}}.$

Therefore, $St_{DNP}\frac{1}{Z_{m}}\rightarrow\frac{1}{z}.$ Hence, we write
$St_{DNP}\frac{Y_{m}}{Z_{m}}\rightarrow\frac{y}{z},\;z\neq 0.$

  5. (5)

Suppose that $St_{DNP}Y_{m}\rightarrow Y$ and $X$ be a random variable such
that $Y_{m}X\rightarrow YX.$ Since $X$ is a random variable such that
$\forall\varepsilon>0,\;\exists\;\delta>0$ and
$e_{y_{m}-n}g_{n}P(|X|>\delta)\leq\frac{\varepsilon}{2}.$ Next,
$\forall\;\varepsilon^{\prime}>0,$

$\displaystyle e_{y_{m}-n}g_{n}P\big{(}|Y_{m}X-YX|\geq\varepsilon^{\prime})$
$\displaystyle=$ $\displaystyle
e_{y_{m}-n}g_{n}P\big{(}|Y_{m}-Y||X|\geq\varepsilon^{\prime},|Z|>\delta\big{)}$
$\displaystyle+e_{y_{m}-n}g_{n}P\big{(}|Y_{m}-Y||X|\geq\varepsilon^{\prime},|Z|\leq\delta\big{)}\leq\frac{\varepsilon}{2}$
$\displaystyle+e_{y_{m}-n}g_{n}P\big{(}|Y_{m}-Y|\geq\frac{\varepsilon^{\prime}}{\delta}\big{)}.$

Which implies,
$\big{\\{}n:n\leq\mathcal{R}_{m}\;\mbox{and}\;e_{y_{m}-n}g_{n}P\big{(}|Y_{m}X-YX|\geq\varepsilon^{\prime}\big{)}\big{\\}}\subseteq\big{\\{}n:n\leq\mathcal{R}_{m}\;\mbox{and}\;e_{y_{m}-n}g_{n}P\big{(}|Y_{m}-y|\geq\frac{\varepsilon^{\prime}}{\delta}\big{)}\geq\frac{\varepsilon}{2}\big{\\}}.$
Therefore,

$St_{DNP}(Y_{m}-Y)(Z_{m}-Z)\rightarrow 0.$

Thus,

$St_{DNP}Y_{m}Z_{m}\rightarrow YZ.$

  6. (6)

Suppose that $(x_{m})$ and $(y_{m})$ be two non-negative sequences such that

$e_{y_{m}-n}g_{n}P\Big{(}|Y_{m}-Y|\geq\frac{\varepsilon}{2}\Big{)}<\frac{\delta}{2}$

and

$\Big{\\{}n:n\leq\mathcal{R}_{m}\;\mbox{and}\;e_{y_{m}-n}g_{n}P\Big{(}|Y_{m}-Y|\geq\frac{\varepsilon}{2}\Big{)}<\frac{\delta}{2}\Big{\\}}=1.$

Now,
$\big{\\{}n:n\leq\mathcal{R}_{m}\;\mbox{and}\;e_{y_{m}-n}g_{n}P\big{(}|Y_{m}-Y|\geq\varepsilon\big{)}\geq\delta\big{\\}}\subseteq\big{\\{}n:n\leq\mathcal{R}_{m}\;\mbox{and}\;e_{y_{m}-n}g_{n}P\big{(}|Y_{m}-Y|\geq\frac{\varepsilon}{2}\big{)}<\frac{\delta}{2}\big{\\}}=1.$
Which implies that

$d(\\{n:n\leq\mathcal{R}_{m}\;\mbox{and}\;e_{y_{m}-n}g_{n}P(|Y_{m}-Y|\geq\varepsilon)\geq\delta\\})=0.$

∎

######  Theorem 2.2.

Suppose that $f:\mathbb{R}\rightarrow\mathbb{R}$ is uniform continuous on
$\mathbb{R}$ and $St_{DNP}Y_{m}\rightarrow Y.$ Then
$St_{DNP}f(Y_{m})\rightarrow f(Y).$

###### Proof.

Let us consider a random variable $Y$ such that for each
$\delta>0,\;\exists\;\beta\in\mathbb{R}$ such that $P(Y>\beta)\leq\delta/2.$
Since, $f$ is uniformly continuous on
$[\beta,\beta]\;\forall\varepsilon>0,\;\exists\delta_{0}$ such that

$|f(y_{m})-f(y)|<\varepsilon\;\mbox{whenever}\;|y_{m}-y|<\delta_{0}.$

Thus,

$\displaystyle P(|f(Y_{m})-f(Y)|\geq\varepsilon)$ $\displaystyle\leq$
$\displaystyle P(|Y_{m}-Y|\geq\delta_{0})+P(|Y>\beta|)$ $\displaystyle\leq$
$\displaystyle P(|Y_{m}-Y|\geq\delta_{0})+\delta/2.$

However, from the definition of $St_{DNP}-$convergence, we have

$\Big{\\{}n:n\leq\mathcal{R}_{m}\;\mbox{and}\;e_{y_{m}-n}g_{n}P\Big{(}|f(Y_{m})-f(Y)|\geq\varepsilon\Big{)}\geq\delta\Big{\\}}$
$\displaystyle\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\subseteq\Big{\\{}n:n\leq\mathcal{R}_{m}\;\mbox{and}\;e_{y_{m}-n}g_{n}P\Big{(}|Y_{m}-Y|\geq\delta_{0}\Big{)}<\frac{\delta}{2}\Big{\\}}.$

∎

##  3\. Deferred Nörlund statistical mean convergence

######  Definition 3.1.

Suppose that $r\geq 1$ be a fixed number. A sequence $(Y_{m})$ is $r^{th}$
mean convergent to $Y,$ if

$\lim_{m\rightarrow\infty}E(|Y_{m}-Y|^{r})=0.$

######  Definition 3.2.

A sequence $(Y_{m})$ is statistically $r^{th}$ mean convergent $(MC)$ to a
random variable $Y,$ where $Y:S\rightarrow\mathbb{R}$ if,

$\lim_{m\rightarrow\infty}\frac{1}{m}\Big{|}n:n\leq
m\;\;\mbox{and}\;\;E(|Y_{m}-Y|^{r}\geq\varepsilon)\Big{|}=0$

for any $\varepsilon>0.$
We write it as

$St_{MC}\lim_{m\rightarrow\infty}E(|Y_{m}-Y|^{r})=0.$

######  Definition 3.3.

Suppose that $(x_{m})$ and $(y_{m})$ are the sequences fulfilling conditions
(1.1) and $(e_{m}),(g_{m})$ are sequences satisfying (1.2). A sequence
$(Y_{m})$ is said to be deferred Nörlund statistically $r^{th}\;(r\geq 1)$
mean convergent to $Y$ ($Y:S\rightarrow\mathbb{R}),$ if for $\varepsilon>0,$

$\displaystyle\lim_{m\rightarrow\infty}\frac{1}{\mathcal{R}_{m}}\Big{|}\Big{\\{}n:n\leq\mathcal{R}_{m}\;\;\mbox{and}\;\;e_{y_{m}-n}g_{n}E\big{(}|Y_{m}-Y|^{r}\geq\varepsilon\big{)}\Big{\\}}\Big{|}=0.$

It is denoted as

$St_{DNM}\lim_{m\rightarrow\infty}E(|Y_{m}-Y|^{r})=0.$

######  Theorem 3.4.

Let $St_{DNM}\lim_{m\rightarrow\infty}E(|Y_{m}-Y|^{r})=0\;\mbox{for}\;r\geq
1,$ then $St_{DNM}\lim_{m\rightarrow\infty}P(|Y_{m}-Y|\geq\varepsilon)=0.$

###### Proof.

For every $\varepsilon>0,$ we have from Markov’s inequality

$\displaystyle St_{DNM}\lim_{m\rightarrow\infty}P(|Y_{m}-Y|\geq\varepsilon)$
$\displaystyle=$ $\displaystyle
St_{DNM}\lim_{m\rightarrow\infty}P(|Y_{m}-Y|^{r}\geq\varepsilon^{r})\;\;(r\geq
1)$ $\displaystyle\leq$ $\displaystyle
St_{DNM}\lim_{m\rightarrow\infty}\frac{E(|Y_{m}-Y|^{r})}{\varepsilon^{r}}=0.$

From definition of statistically deferred Nörlund mean convergence

$St_{DNM}\lim_{m\rightarrow\infty}E(|Y_{m}-Y|^{r})=0,$

it implies that

$St_{DNP}\lim_{m\rightarrow\infty}P(|Y_{m}-Y|\geq\varepsilon)=0.$

∎

We now present an example to show that a sequence of random variables is
statistically probability convergent but not statistically $r^{th}-$mean
convergent.

Example 1: Suppose that $x_{m}=2m-1,y_{m}=4m-1.$ Also, suppose that
$e_{y_{m}-m}=2m$ and $g_{m}=1.$ Further, consider a sequence $(z_{m})$ of
random variables such that

$Y_{m}=\left\\{\begin{array}[]{ll}m,\;\mbox{with
probability}\;\frac{1}{\sqrt{m}}\\\ \\\ 0,\;\mbox{with
probability}\;1-\frac{1}{\sqrt{m}}.\\\ \end{array}\right.$

Then the statistically deferred Nörlund convergence of $Y_{m}$ is given as

$\displaystyle\lim_{m\rightarrow\infty}\frac{1}{2m}\Big{|}\Big{\\{}n:n\leq\mathcal{R}_{m}\;\;\mbox{and}\;\;2mP(|Y_{m}-0|\geq\varepsilon)\Big{\\}}\Big{|}$
$\displaystyle=$ $\displaystyle\lim_{m\rightarrow\infty}P(Y_{m}=m)$
$\displaystyle=$ $\displaystyle\lim_{m\rightarrow\infty}\frac{1}{\sqrt{m}}$
$\displaystyle=$ $\displaystyle 0.$

However, statistically deferred Nörlund mean convergence, for $r\geq 1,$ is

$\displaystyle\lim_{m\rightarrow\infty}\frac{1}{2m}\Big{|}\Big{\\{}n:n\leq\mathcal{R}_{m}\;\;\mbox{and}\;\;2mE(|Y_{m}-0|^{r})\Big{\\}}\Big{|}$
$\displaystyle=$
$\displaystyle\lim_{m\rightarrow\infty}\Big{(}m^{r}\Big{(}\frac{1}{\sqrt{m}}\Big{)}+0\Big{(}1-\frac{1}{\sqrt{m}}\Big{)}\Big{)}$
$\displaystyle=$ $\displaystyle\lim_{m\rightarrow\infty}m^{r-1/2}$
$\displaystyle=$ $\displaystyle\infty.$

This implies that the sequence $(Y_{m})$ is $St_{DNP}-$convergent but not
$St_{DNM}-$convergent.

##  4\. Statistical distribution convergence via Deferred Nörlund

######  Definition 4.1.

The sequence of random variables $(Y_{m})$ is said to be distribution
convergent (or convergent in distribution) to $Y,$ if

$\lim_{m\rightarrow\infty}F_{Y_{m}}(y)=F_{Y}(y)$

for all $y\in\mathbb{R}$ at which $F_{Y}(y)$ is continuous.

Thorughout the paper $(F_{Y_{m}}(y))$ is the sequence of distribution
functions of $(Y_{m})$ and $F_{Y}(y)$ is the distribution function of $Y.$

######  Definition 4.2.

The sequence $(F_{Y_{m}}(y))$ is called as statistically distribution
convergent $(or\;St_{DC}),$ if there exists $F_{Y}(y)$ of random variable $Y$
such that for each $\varepsilon>0,$

$\displaystyle\lim_{m\rightarrow\infty}\frac{1}{m}\Big{|}\Big{\\{}n:n\leq
m\;\;\mbox{and}\;\;|F_{Y_{m}}(y)-F_{Y}(y)|\geq\varepsilon\Big{\\}}\Big{|}=0.$

We may write this as

$St_{DC}\lim_{m\rightarrow\infty}F_{Y_{m}}(y)=F_{Y}(y).$

######  Definition 4.3.

The sequence $(F_{Y_{m}}(y))$ of distribution functions is called as deferred
Nörlund statistically distribution convergent $(or\;St_{DNDC}),$ if there
exists $F_{Y}(y)$ of $Y$ such that for each $\varepsilon>0,$

$\displaystyle\lim_{m\rightarrow\infty}\frac{1}{\mathcal{R}_{m}}\Big{|}\Big{\\{}n:n\leq\mathcal{R}_{m}\;\;\mbox{and}\;\;e_{y_{m}-n}g_{n}|F_{Y_{m}}(y)-F_{Y}(y)|\geq\varepsilon\Big{\\}}\Big{|}=0.$

In this case, we say

$\displaystyle St_{DNDC}\lim_{m\rightarrow\infty}F_{Y_{m}}(y)$
$\displaystyle=$ $\displaystyle F_{Y}(y).$

######  Theorem 4.4.

Suppose that $St_{DNP}\lim_{m\rightarrow\infty}P(|Y_{m}-Y|\geq\varepsilon)=0,$
then

$St_{DNDC}\lim_{m\rightarrow\infty}F_{Y_{m}}(y)=F_{Y}(y).$

###### Proof.

Suppose that $(F_{Y_{m}}(y))$ is distribution functions of $(Y_{m}),$ and
$F_{Y}(y)$ be the distribution function of $Y.$ For $i,j\in\mathbb{R}$ such
that $i<j,$ we have

$(Y\leq i)=(Y_{m}\leq j,Y\leq i)+(Y_{m}\geq j,Y\leq i).$

Further,

$(Y_{m}\leq j,Y\leq i)\subseteq(Y_{m}\leq j),$

which implies that

(4.1) $(Y\leq i)\subseteq(Y_{m}\leq j)+(Y_{m}\geq j,Y\leq i).$

Let us take the probability to left hand side and right hand side of equation
(4.1)

$\displaystyle P(Y\leq i)$ $\displaystyle\leq$ $\displaystyle P\\{(Y_{m}\leq
j)+(Y_{m}\geq j,Y\leq i)\\}$ $\displaystyle\leq$ $\displaystyle P(Y_{m}\leq
j)+P(Y_{m}\geq j,Y\leq i).$

It means that

(4.2) $F_{Y_{m}}(j)\geq F_{Y}(i)-P(Y_{m}\geq j,Y\leq i).$

If $Y_{m}\geq j,Y\leq i,$ then $Y_{m}\geq j,-Y\geq-i,$ so that $Y_{m}-Y>j-i,$
that is,

$(Y_{m}\geq j,Y\leq i)\subseteq(Y_{m}-Y>j-i)\subseteq(|Y_{m}-Y|>j-i).$

This means

$P(Y_{m}\geq j,Y\leq i)\leq P(|Y_{m}-Y|>j-i).$

As we know that $i<j$ and $St_{DNP}Y_{m}\rightarrow Y,$ we obtain

$St_{DNP}\lim_{m\rightarrow\infty}P(Y_{m}\geq j,Y\leq i)=0.$

From (4.2) we get

$St_{DNDC}\lim_{m\rightarrow\infty}F_{Y_{m}}(j)\geq F_{Y}(i).$

Similarly, if $j<a$ for any real constant $a,$ then

$(Y\leq j)=(Y\leq a,Y_{m}\leq j)+(Y>a,Y_{m}\leq j).$

Consequently,

$F_{Y_{m}}(j)\leq F_{Y}(a)+P(Y>a,Y_{m}\leq j)$

and

$St_{DNDC}\lim_{m\rightarrow\infty}P(Y>a,Y_{m}\leq j)=0.$

Therefore, we get

$St_{DNDC}\lim_{m\rightarrow\infty}F_{Y_{m}}(j)\leq F_{Y}(a).$

Thus, with $i<j<a,$ we have

$St_{DNDC}\lim_{m\rightarrow\infty}F_{Y_{m}}(j)=F_{Y}(i).$

∎

Example 2: Consider the random variables $((Y_{m}),Y)$ of two dimensions as
$\\{(0,0),(0,1),\\\ (1,0),(1,1)\\}$ such that

$(Y_{m},Y)=\left\\{\begin{array}[]{ll}0,\;\;[P(Y_{m}=0,Y=0)=0=P(Y_{m}=1,Y=1)]\\\
\\\ \frac{1}{2},\;[P(Y_{m}=1,Y=0)=0=P(Y_{m}=0,Y=1)].\\\ \\\
\end{array}\right.$

The distribution function of $Y_{m}$ is given by $Y_{m}=(\lambda_{1}=0,1),$
with probability mass function

$(p_{y_{m},\lambda_{1}})=P(Y_{m}=\lambda_{1}),\;\mbox{where}\;p_{y_{m},0}=\frac{1}{2}=p_{y_{m},1}$

and for $Y=\lambda_{2}(\lambda_{2}=0,1),$ with probability mass function

$(p_{y_{m},\lambda_{2}})=P(Y_{m}=\lambda_{2}),\;\mbox{where}\;p_{y,0}=\frac{1}{2}=p_{y,1}.$

If $(F_{Y_{m}}(y))$ is distribution functions of $(Y_{m})$ and $F_{Y}(y)$ is
the distribution function of $Y,$ then

$F_{Y}(y)=\lim_{m\rightarrow\infty}F_{Y_{m}}(y)=\left\\{\begin{array}[]{ll}0,\;\;(y<0)\\\
\\\ \frac{1}{2},\;\;(0\leq y<1).\\\ \\\ 1,\;\;(z>1).\end{array}\right.$

Thus, we get

$St_{DNDC}\lim_{m\rightarrow\infty}F_{Y_{m}}(y)=F_{Y}(y),\;\mbox{where}\;x_{m}=2m-1,y_{m}=4m-1,e_{y_{m}-m}=2m\;\mbox{and}\;g_{m}=1.$

But, it is not $St_{PC}$ for the sequence of random variables, i.e.

$St_{DNPC}\lim_{m\rightarrow\infty}P(|Y_{m}-Y|\geq\varepsilon)\neq
0,\;\mbox{where}\;x_{m}=2m-1,y_{m}=4m-1,e_{y_{m}-m}=2m\;\mbox{and}\;g_{m}=1.$

##  5\. Applications

The hypothesis of the Korovkin-type theorems have been studied by several
researchers in various field in different ways such as in, summability theory,
functional analysis and probability theory. Korovkin-type approximation
theorems have been investigated by many mathematicians under various
background, involving function spaces, Banach spaces, and so on. Recently,
Mohiuddine and Alamri studied Korovkin and Voronovskaya type approximation
theorems in [11]. Further, Hazarika et al. [6] studied Korovkin approximation
theorem for Bernstein operator of rough statistical convergence of triple
sequences. For detailed study on Korovkin approximation theorem one may refer
[2], [13], [14], [18], [22].
By $\mathcal{C}(Y),$ we denote the space of all continuous probability
functions defined on a compact subset $Z\subset\mathbb{R}.$ The space
$\mathcal{C}(Y)$ is a Banach space with respect to the norm

$||f||_{\infty}=\displaystyle\sup_{z\in
Y}\\{|f(z)|\\},\;\;\;\;\;f\in\mathcal{C}(Y).$

We say that $\mathcal{Y}$ is a positive linear operator of sequence of random
variables if

$\mathcal{Y}(f,z)\geq 0\;\mbox{whenever}\;f\geq 0.$

Throughout, $\mathcal{Y}_{n}:\mathcal{C}(Y)\rightarrow\mathcal{C}(Y)$ be a
sequence of random variables of positive linear operators.

######  Theorem 5.1.

[26] Let $\mathcal{Y}_{n}:\mathcal{C}(Y)\rightarrow\mathcal{C}(Y)$. Then for
all $f\in\mathcal{C}(Y),$ we have

(5.1)
$St_{DNP}\lim_{n\rightarrow\infty}||\mathcal{Y}_{n}(f,z)-f(z)||_{\infty}=0$

iff

(5.2)
$St_{DNP}\lim_{n\rightarrow\infty}||\mathcal{Y}_{n}(1,z)-1||_{\infty}=0,$
(5.3)
$St_{DNP}\lim_{n\rightarrow\infty}||\mathcal{Y}_{n}(z,z)-z||_{\infty}=0,$
(5.4)
$St_{DNP}\lim_{n\rightarrow\infty}||\mathcal{Y}_{n}(z^{2},z)-z^{2}||_{\infty}=0.\\\
$

######  Theorem 5.2.

Let $\mathcal{Y}_{n}:\mathcal{C}(Y)\rightarrow\mathcal{C}(Y)$. Then for all
$f\in\mathcal{C}(Y),$ we have

(5.5)
$St_{DNM}\lim_{n\rightarrow\infty}||\mathcal{Y}_{n}(f,z)-f(z)||_{\infty}=0$

iff

(5.6)
$St_{DNM}\lim_{n\rightarrow\infty}||\mathcal{Y}_{n}(1,z)-1||_{\infty}=0,$
(5.7)
$St_{DNM}\lim_{n\rightarrow\infty}||\mathcal{Y}_{n}(z,z)-z||_{\infty}=0,$
(5.8)
$St_{DNM}\lim_{n\rightarrow\infty}||\mathcal{Y}_{n}(z^{2},z)-z^{2}||_{\infty}=0.\\\
$

######  Theorem 5.3.

Let $\mathcal{Y}_{n}:\mathcal{C}(Y)\rightarrow\mathcal{C}(Y)$. Then for all
$f\in\mathcal{C}(Y),$ we have

(5.9)
$St_{DNDC}\lim_{n\rightarrow\infty}||\mathcal{Y}_{n}(f,z)-f(z)||_{\infty}=0$

iff

(5.10)
$St_{DNDC}\lim_{n\rightarrow\infty}||\mathcal{Y}_{n}(1,z)-1||_{\infty}=0,$
(5.11)
$St_{DNDC}\lim_{n\rightarrow\infty}||\mathcal{Y}_{n}(z,z)-z||_{\infty}=0,$
(5.12)
$St_{DNDC}\lim_{n\rightarrow\infty}||\mathcal{Y}_{n}(z^{2},z)-z^{2}||_{\infty}=0.\\\
$

Example 3: Let $\mathcal{M}_{m}(f,y)$ be a Meyer-König and Zeller operators on
$\mathcal{C}[0,1]$ and $Z=[0,1]$ as defined in [1] as

$\mathcal{M}_{m}(f,y)=(1-y)^{m+1}\sum_{t=0}^{\infty}f\Big{(}\frac{t}{t+m+1}\Big{)}\binom{m+t}{t}y^{t}.$

Further, let us consider a sequence of operators
$\mathcal{Y}_{n}:\mathcal{C}[0,1]\rightarrow\mathcal{C}[0,1]$ and $(Y_{n})$ as
defined in example $2.4$ such that

(5.13)
$\mathcal{Y}_{n}(f,y)=[1+F_{Y_{m}}(y)]\mathcal{M}_{n}(f),\;\;\;(f\in\mathcal{C}[0,1]),\\\
$

where $(F_{Y_{m}}(y))$ is defined in Example 2. Now we observe that

$\mathcal{Y}_{n}(1,y)=[1+F_{Y_{m}}(y)]\cdot 1=[1+F_{Y_{m}}(y)],$
$\mathcal{Y}_{n}(u,z)=[1+F_{Y_{m}}(y)]\cdot y=[1+F_{Y_{m}}(y)]\cdot y$

and

$\displaystyle\mathcal{Y}_{n}(u^{2},z)$ $\displaystyle=$
$\displaystyle[1+F_{Y_{m}}(y)]\cdot\Big{\\{}y^{2}\Big{(}\frac{m+2}{m+1}\Big{)}+\frac{y}{m+1}\Big{\\}}.$

Therefore, we have

$\displaystyle
St_{DNDC}\lim_{n\rightarrow\infty}||\mathcal{Y}_{n}(1,y)-1||_{\infty}=0,$
$\displaystyle
St_{DNDC}\lim_{n\rightarrow\infty}||\mathcal{Y}_{n}(y,y)-y||_{\infty}=0,$
$\displaystyle
St_{DNDC}\lim_{n\rightarrow\infty}||\mathcal{Y}_{n}(y^{2},y)-y^{2}||_{\infty}=0.$

Hence, $\mathcal{Y}_{n}(f,y)$ fulfills (5.10), (5.11) and (5.12). Thus, from
Theorem 5.3

$St_{DNDC}\lim_{n\rightarrow\infty}||\mathcal{Y}_{n}(f,y)-f||_{\infty}=0.$

Hence, it is $(DNDC)-$convergent. However, $(Y_{m})$ is neither
$(DN)-$statistical convergent nor $(DN)-$convergent. Thus, we can exhibit that
the work in [24] does not hold for our operators described in (5.13). Hence,
our Theorem 5.3 is stronger than the theorem proved in [24].

##  6\. Conclusion

Upon prior analysis, our interest is to modify the studies of Srivastava et
al. [25] and introduce various aspects of statistical convergence for the
sequences of random variables and sequences of real numbers via deferred
Norlund summability mean. We first study various results presenting the
connection by using fundamental limit concepts of sequences of random
variables. As an applications of our findings, we present new Krorvkin-type
approximation results and also demonstrated the effectiveness of the findings.
As a future work one can obtain the corresponding results of the present paper
using deferred Euler summability mean.

Compliance with ethical standards

Availability of data and material: Not applicable.

Conflict of interest: The authors declare that they have no conflict of
interest.

Ethical approval: This article does not contain any studies with human
participants or animals performed by any of the authors.

Acknowledgement: The corresponding author thanks the Council of Scientific and
Industrial Research (CSIR), India for partial support under Grant No.
25(0288)/18/EMR-II, dated 24/05/2018.

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