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Asked By :  Carlos
Answers1

Suppose x1 poisson1 and x2 poisson2 are independent what is

Suppose X1 ∼ Poisson(λ1) and X2 ∼ Poisson(λ2) are independent. What is the

distribution of X1 + X2? Find the conditional probability P[X1 = m |X1 + X2 = n]

Here m ≤ n.

Let X1 ∼ Geometric(p1) and X2 ∼ Geometric(p2). Let Y = min(X1, X2) what is

the distribution of Y? Find P[X1 = X2]. Let Z = max(X1, X2) − Y. Find the joint

p.m.f. of (Y, Z).




Answers :

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Part 1: Distribution of X1+X2X_1 + X_2 and Conditional Probability

Given that X1Poisson(λ1)X_1 \sim \text{Poisson}(\lambda_1) and X2Poisson(λ2)X_2 \sim \text{Poisson}(\lambda_2), and they are independent:

Distribution of X1+X2X_1 + X_2:

The sum of two independent Poisson random variables is also Poisson-distributed, with the rate parameter being the sum of their rate parameters:

X1+X2Poisson(λ1+λ2).X_1 + X_2 \sim \text{Poisson}(\lambda_1 + \lambda_2).

Conditional Probability P[X1=mX1+X2=n]P[X_1 = m \mid X_1 + X_2 = n]:

Given X1+X2=nX_1 + X_2 = n, the conditional distribution of X1X_1 is Binomial with parameters nn and λ1λ1+λ2\frac{\lambda_1}{\lambda_1 + \lambda_2}. That is:

P[X1=mX1+X2=n]=(nm)(λ1λ1+λ2)m(λ2λ1+λ2)nm,P[X_1 = m \mid X_1 + X_2 = n] = \binom{n}{m} \left(\frac{\lambda_1}{\lambda_1 + \lambda_2}\right)^m \left(\frac{\lambda_2}{\lambda_1 + \lambda_2}\right)^{n-m},

for m=0,1,,nm = 0, 1, \dots, n.


Part 2: Distribution of Y=min(X1,X2)Y = \min(X_1, X_2)

X1Geometric(p1)X_1 \sim \text{Geometric}(p_1), X2Geometric(p2)X_2 \sim \text{Geometric}(p_2):

The pmf of a geometric random variable XGeometric(p)X \sim \text{Geometric}(p) is:

P(X=k)=(1p)k1p,k=1,2,3,P(X = k) = (1-p)^{k-1}p, \quad k = 1, 2, 3, \dots

Distribution of Y=min(X1,X2)Y = \min(X_1, X_2):

For Y=min(X1,X2)Y = \min(X_1, X_2), the event {Yk}\{Y \geq k\} occurs if both X1kX_1 \geq k and X2kX_2 \geq k. Since X1X_1 and X2X_2 are independent:

P(Yk)=P(X1k)P(X2k).P(Y \geq k) = P(X_1 \geq k) \cdot P(X_2 \geq k).

For geometric random variables:

P(Xik)=(1pi)k1,i=1,2.P(X_i \geq k) = (1 - p_i)^{k-1}, \quad i = 1, 2.

Thus:

P(Yk)=(1p1)k1(1p2)k1=[(1p1)(1p2)]k1.P(Y \geq k) = (1-p_1)^{k-1}(1-p_2)^{k-1} = [(1-p_1)(1-p_2)]^{k-1}.

The pmf of YY is:

P(Y=k)=P(Yk)P(Yk+1),P(Y = k) = P(Y \geq k) - P(Y \geq k+1), P(Y=k)=[(1p1)(1p2)]k1[1(1p1)(1p2)].P(Y = k) = [(1-p_1)(1-p_2)]^{k-1} \cdot [1 - (1-p_1)(1-p_2)].

Hence, YGeometric(p1+p2p1p2)Y \sim \text{Geometric}(p_1 + p_2 - p_1p_2).


Part 3: Probability P[X1=X2]P[X_1 = X_2]

The event X1=X2=kX_1 = X_2 = k occurs with probability:

P(X1=k,X2=k)=P(X1=k)P(X2=k),P(X_1 = k, X_2 = k) = P(X_1 = k) \cdot P(X_2 = k),

since X1X_1 and X2X_2 are independent:

P(X1=k)=(1p1)k1p1,P(X2=k)=(1p2)k1p2.P(X_1 = k) = (1-p_1)^{k-1}p_1, \quad P(X_2 = k) = (1-p_2)^{k-1}p_2.

Thus:

P(X1=X2)=k=1P(X1=k,X2=k),P(X_1 = X_2) = \sum_{k=1}^\infty P(X_1 = k, X_2 = k), P(X1=X2)=k=1[(1p1)(1p2)]k1p1p2.P(X_1 = X_2) = \sum_{k=1}^\infty [(1-p_1)(1-p_2)]^{k-1} p_1 p_2.

This is a geometric series with ratio (1p1)(1p2)(1-p_1)(1-p_2), so:

P(X1=X2)=p1p21(1p1)(1p2)=p1p2p1+p2p1p2.P(X_1 = X_2) = \frac{p_1 p_2}{1 - (1-p_1)(1-p_2)} = \frac{p_1 p_2}{p_1 + p_2 - p_1p_2}.

Part 4: Joint pmf of (Y,Z)(Y, Z)

Let Z=max(X1,X2)YZ = \max(X_1, X_2) - Y. Then Z=X1X2Z = |X_1 - X_2|, and YY represents the smaller value of X1X_1 and X2X_2.

Joint pmf P(Y=y,Z=z)P(Y = y, Z = z):

For Y=yY = y and Z=zZ = z, we have:

  • X1=yX_1 = y, X2=y+zX_2 = y+z, or
  • X2=yX_2 = y, X1=y+zX_1 = y+z.

The joint pmf is:

P(Y=y,Z=z)=P(X1=y,X2=y+z)+P(X2=y,X1=y+z).P(Y = y, Z = z) = P(X_1 = y, X_2 = y+z) + P(X_2 = y, X_1 = y+z).

Using the pmf of geometric distributions:

P(X1=y,X2=y+z)=P(X1=y)P(X2=y+z),P(X_1 = y, X_2 = y+z) = P(X_1 = y)P(X_2 = y+z), P(X1=y)=(1p1)y1p1,P(X2=y+z)=(1p2)y+z1p2.P(X_1 = y) = (1-p_1)^{y-1}p_1, \quad P(X_2 = y+z) = (1-p_2)^{y+z-1}p_2.

Similarly, for the second term:

P(X2=y,X1=y+z)=P(X2=y)P(X1=y+z).P(X_2 = y, X_1 = y+z) = P(X_2 = y)P(X_1 = y+z).

Thus:

P(Y=y,Z=z)=[(1p1)y1p1(1p2)y+z1p2]+[(1p2)y1p2(1p1)y+z1p1].P(Y = y, Z = z) = [(1-p_1)^{y-1}p_1 (1-p_2)^{y+z-1}p_2] + [(1-p_2)^{y-1}p_2 (1-p_1)^{y+z-1}p_1].

Factoring terms:

P(Y=y,Z=z)=(1p1)y1(1p2)y1[(1p2)zp1p2+(1p1)zp2p1].P(Y = y, Z = z) = (1-p_1)^{y-1}(1-p_2)^{y-1} [(1-p_2)^z p_1 p_2 + (1-p_1)^z p_2 p_1].


Answered By

Sasa

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