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A random variable is a real-valued function defined over a sample space. Consequently, a random variable can be used to identify numerical events of interest in an experiment. Random variables may be either continuous or discrete.
A random variable
A set of elements is said to be countably infinite* if the elements in the set can be put into one-to-one correspondence with the positive integers.
Because certain types of random variables occur so frequently in practice, it is useful to have at hand the probability of each value of ra random variable. This collection of probabilities is called the probability distribution of the discrete random variable.
Example: Binomial distribution function,
On the other hand, a random variable
Unfortunately, the probability distribution for a continuous random variable cannot be specified in the same way as outlined above for a discrete random variable; it is mathematically impossible to assign nonzero probabilities to all points on a line interval while satisfying the requirement that the probabilities of the distinct possible values sum to one.
Rather, we define a probability density function for the random variable:
Let
#F(y)# be the distribution function for a continuous random variable#Y# . Then#f(y)# , given by
#f(y)=(dF(y))/(dy)=F'(y)# wherever the derivative exists, is called the probability density function for the random variable
#Y# .
Example: A distribution function
Recognize that
As you stated, for a continuous random variable
The moment generating function is
(a) Therefore,
(b) The probability is
(c) In the general case, the mean is
The mean is also the "first moment"
The second moment is
The variance
To determine the probability of an event happening, there are two values you need to determine first.
How many possible outcomes are there altogether?
In this case with a toss of the die, there are
Of the possible outcomes how many of them will meet our conditions? IN this case we want a number equal to or less than
Write a probability as a fraction and give in its simplest form.
Given
Let
Let
The support is the 2-D region
a)
When
#E[g(X,Y)] = int_A int_B g(x,y)f_(X,Y)(x,y)" "dy" "dx#
Of course, the function
#E[X] = int_A int_B x" "f_(X,Y)(x,y)" "dy" "dx#
For this question:
#E(X) = int_(x=0)^1int_(y=0)^1x(4x^2y+2y^5)" "dy" "dx#
#color(white)(E(X)) = int_(0)^1 x int_(0)^1(4x^2y+2y^5)" "dy" "dx#
#color(white)(E(X)) = int_(0)^1 x [2x^2y^2+1/3 y^6]_(y=0)^1" "dx#
#color(white)(E(X)) = int_(0)^1 x (2x^2+1/3)" "dx#
#color(white)(E(X)) = int_(0)^1 (2x^3+1/3 x)" "dx#
#color(white)(E(X)) = [1/2 x^4+1/6 x^2]_(x=0)^1#
#color(white)(E(X)) = 1/2 + 1/6#
#color(white)(E(X)) = 2/3#
b)
I'll leave the calculation of
c)
Using the rules of expectation, we get
#E(3X+7Y)=E(3X) + E(7Y)#
#color(white)(E(3X+7Y))=3E(X) + 7E(Y)#
#color(white)(E(3X+7Y))=3(2/3) + 7(46/63)#
#color(white)(E(3X+7Y))=2+46/9#
#color(white)(E(3X+7Y))=64/9#
d)
Recall by the definition of variance:
#"Var"(X) = E{[X-E(X)]^2} = E(X^2)- [E(X)]^2#
We already know
#E(X^2) = int_(x=0)^1int_(y=0)^1x^2(4x^2y+2y^5)" "dy" "dx#
#color(white)(E(X^2)) = int_(0)^1 x^2 int_(0)^1(4x^2y+2y^5)" "dy" "dx#
#color(white)(E(X^2)) = int_(0)^1 x^2 [2x^2y^2+1/3 y^6]_(y=0)^1" "dx#
#color(white)(E(X^2)) = int_(0)^1 x^2 (2x^2+1/3)" "dx#
#color(white)(E(X^2)) = int_(0)^1 (2x^4+1/3 x^2)" "dx#
#color(white)(E(X^2)) = [2/5 x^5+1/9 x^3]_(x=0)^1#
#color(white)(E(X^2)) = 2/5 + 1/9#
#color(white)(E(X^2)) = 23/45#
Thus,
#"Var"(X) = E(X^2)- [E(X)]^2#
#color(white)("Var"(X)) = 23/45- (2/3)^2#
#color(white)("Var"(X)) = 23/45- 4/9#
#color(white)("Var"(X)) = 1/15#
e)
Once again, I'll leave this one as practice. You should get
f)
Same as before, but with
g)
We use the properties of expectation again to get
#E(XY+14)=E(XY) + E(14)#
#color(white)(E(XY+14))=10/21 + 14#
#color(white)(E(XY+14))=304/21#
h)
We're looking for the probability of being in the part of the support that satisfies
#x+y<=1#
#=>" "y<=1-x#
Combining this with the given restrictions
We can picture covering this region as follows: as
Thus, our probability is
#"P"(X+Y<=1)= int_(x=0)^1int_(y=0)^(1-x)f_(X,Y)(x,y)" "dy" "dx#
#color(white)("P"(X+Y<=1))= int_(0)^1int_(0)^(1-x)(4x^2y+2y^5)" "dy" "dx#
#color(white)("P"(X+Y<=1))= int_(0)^1 [2x^2y^2]_(y=0)^(1-x)" "dx#
#color(white)("P"(X+Y<=1))= int_(0)^1 2x^2(1-x)^2" "dx#
#color(white)("P"(X+Y<=1))= int_(0)^1 (2x^2-4x^3+2x^4)" "dx#
#color(white)("P"(X+Y<=1))= [2/3x^3-x^4+2/5x^5]_(x=0)^1#
#color(white)("P"(X+Y<=1))= 2/3-1+2/5#
#color(white)("P"(X+Y<=1))= 1/15#
To find the variance, we first need to calculate the mean.
To calculate the mean, simply add all the data points, then divide by the number of data points.
The formula for the mean
Where
For our data set, we have:
So the mean is
Now to calculate the variance, we find out how far away each data point is from the mean, then square each of those values, add them up, and divide by the number of data points.
The variance is given the symbol
The formula for the variance is:
So for our data:
Let,
and a Black-coloured, resp.
Then, the Reqd. Prob. is,
We know that,
For P(A), there are
selected in
There are
Similarly,
Note that,
an ace and black-coloured, i.e., a black-coloured ace.
Out of
Clearly,
From
derived by Respected Andrea S. !
Enjoy Maths.!
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